346 + Interview Questions in MOTION-IN-A-STRAIGHT-LINE IN GENERAL AWARENESS Page 7 InterviewSolution

301.	A particle starting from rest. Its acceleration (a) versus time (t) is as shown in the figure. The maximum speed of the particle will be. A. `110 ms^(-1)`B. `55 ms^(-1)`C. `550 ms^(-1)`D. `660 ms^(-1)`
Answer» Correct Answer - B Till 11s, acceleration is positive, so velocity will go on increaseing up to 11 s and maximum velocity will happened at 11s. The area under the acceleration -time graph gives change in velocity. Since particle starts with `u = 0`, change in velocity is `v_(f) - v_(i) = v_("max") - 0 =` Area under `a - t` graph or `v_("max") = (1)/(2) xx 10 xx 11 = 55 ms^(-1)`

Discussion

302.	A particle starting from rest. Its acceleration (a) versus time (t) is as shown in the figure. The maximum speed of the particle will be. A. 110 m/sB. 55 m/sC. 550 m/sD. 660 m/s
Answer» Correct Answer - B The velocity increase upto `t = 11s` as acceleration remains positive (although it decrease). `:.` Max. velocity = Area under `v - t` graph `= (1)/(2) xx 11 xx 10 = 55 m//s`

Discussion

303.	A particle starting from rest. Its acceleration (a) versus time (t) is as shown in the figure. The maximum speed of the particle will be. A. `110 m//s`B. `55 m//s`C. `550 m//s`D. `660 m//s`
Answer» Correct Answer - B Area of a-t graph`=1/2xx10xx11=55` `v_(t=11)-v_(t=0)=55` `v_(t=11)=55 m//s`

Discussion

304.	A monkey climbs up a slippery pole for ` 3 seconds` and subsequently slips for `3 seconda`. Its velocity at time (t) is given by ` v (t) =2 t (3-t) , 0 lt tgt 3 s and v (t) =- (t-30 (6-t) for 3 lt tlt 6 s in m//s`. It repeats thei cycle till it reaches the height of ` 20 s.` (a) AT wht time is its v elocity maximum ? (b) At what time is its average velocity maximum ? (c ) At what time is its accelration maximum in magnitude ? (d) How many cycles (counting fractions ) are required to reach the top ?
Answer» It this problem to calculate maximum velocity we will use `(dv)/(dt) =0`, then the time corresponding to maximum velocity will be obntained. Given velocity `v(t) = 2t (3-t) = 6t - 2t^(2)"…….."(i)` (a) For maximum velocity `(dv(t))/(dt) = 0` `rArr d/(dt) (6t - 2t^(2)) = 0` `rArr 6 - 4t =0` `rArr t = (6)/(4) = (3)/(2)s = 15s` (b) From Eq. (i) `v = 6t - 2t^(2)` `rArr (ds)/(dt) = 6t - 2t^(2)` `rArr ds = (6t - 2t^(2))dt` where,s is displacement `:.` Distane travelled in time interval 0 to 3s. `s = int_(0)^(3) (6t - 2t^(2)) dt` `= [(6t^(2))/(2) - (2t^(3))/(3)]_(0)^(3) = [3t^(3) - (2)/(3)t^(3)]_(0)^(3)` `= 3 xx 9 - (2)/(3) xx 3 xx 3 xx 3` `= 27 - 18 = 9 m` Average velocity ` = ("Distance travelled")/("Time")` `= 9/3 = 3 m//s` Given, `x = 6t - 2t^(2)` `rArr 3 = 6t - 2t^(2)` `rArr 2t^(2) - 6t - 3 = 0` `rArr t = (6 +- sqrt(6^(2) - 4 xx 2 xx 3))/(2 xx 2) = (6 +- sqrt(36-24))/(4)` `= (6+- sqrt(12))/(4) = (3+- 2sqrt(3))/(2)` (c) In a periodic motion when velocinty is zero accleration will be maximum putting `v = 0` in Eq. (i) `0 = 6t - 2t^(2)` `rArr 0 = t (6-2t)` `= t xx 2 (3-t) = 0` `rArr t = 0` or `3s` (d) Distance covered in `0` to `3s = 9m` Distance covererd in 3 to 6s = `int_(3)^(6) (18 - 9t +t^(2)) dt` `= (18t - (9t^(2))/(2) + (t^(3))/(3))_(3)^(6)` `= 18 xx 6 - (9)/(2) xx 6^(2) + (6^(3))/(3) - (18 xx 3 - (9 xx 3^(2))/(2) + (3^(3))/(3))` `= 108-9xx18+(6^(3))/(3) - 18 xx 3 + (9)/(2) xx 9 - (27)/(3)` `= 108-18 xx 9 + (216)/(3) - 54+45xx9-9 = - 45 m` `:.` Total distance travelled in one cycle `= s_(1) + s_(2) = 9 - 45 = 45m` Number of cycles covered in total distance to be covered `= (20)/(4.5) = 4.44 ~~ 5`

Discussion

305.	The motion of a particle is described by the equation `x = a+bt^(2)` where `a = 15` cm and `b = 3 cm//s`. Its instantaneous velocity at time 3 sec will beA. `33 cm s^(-1)`B. `18 cm s^(-1)`C. `16 cm s^(-1)`D. `32 cm s^(-1)`
Answer» Correct Answer - A Here, x = `at + bt^(2)` where a = `15 cm s^(-1)`, b = `3 cm s^(-2)` Velocity, `v = dx/dt = d/dt(at + bt^(2)) = a + bt` At t = 3 s, v = `15 + 2 xx 3 xx 3 = 33 cm s^(-1)`

Discussion

306.	Give example of a motion where ` x gt 0, vlt 0, agt ` at a particular instant.
Answer» Let the motion is represented by `x(t) = A + Be^(-gammat)` Let `A gt B` and `gamma gt 0` Now velocity `x(t) = (dx)/(dt) = -Bgammae^(-gammat)` Acceleration `a(t) = (dx)/(dT) = Bgamma^(2)e^(-gammat)` Suppose we are considering any instant t, then from Eq. (i) , we can say that `x(t) gt 0, v(t) lt 0` and `a gt 0`

Discussion

307.	Deduce the equations of unifromly accelerated motion in one dimension by following calculus method.
Answer» By definition `a = (dv)/(dt)` `dv = a dt` Integrating both sides `int_(v_(0))^(v) dv = int_(0)^(t) a dt` `= a int_(0)^(t) dt` (a is constant) `v - v_(0) = at` `v = v_(0) + at` Further, `v = (dx)/(dt)` ltbr. `dx = v dt` Integrating both sides `int_(x_(0))^(x) dx = int_(0)^(t) v dt` `= int_(0)^(t) (v_(0) + at) dt` `x - x_(0) = v_(0) t + (1)/(2) a t^(2)` `x = x_(0) + v_(0) t + (1)/(2) a t^(2)` We can write `a = (dv)/(dt) = (dv)/(dx) (dx)/(dt) = v (dv)/(dx)` or, `v dv = a dx` Integrating both sides `int_(v_(0))^(v) vdv = int_(x_(0))^(x) a dx` `(v^(2) - v_(0)^(2))/(2) = a (x - x_(0))` `v^(2) = v_(0)^(2) + 2a (x - x_(0))` The advantage of this method is that it can be used for motion with non-uniform acceleration also. Now, we shall use these equations to some important cases.

Discussion

308.	Assertion : The accelerated motion of an object may be due to change in magnitude of velocity or direction of velocity or both. Reason : Acceleration can be produed only by change in the magnitude of the velocity.A. If both assertion and reason are trure and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanations of assertion.C. If assertion is true but reason is false.D. If both assertion and reason is false
Answer» Correct Answer - (c ) Change in the direction of the velocity also produces acceleration as in the case of circular motion. The change in the magnitude or direction of the velocity or both can produce acceleration.

Discussion

309.	Assertion : A particle may be momentarily at rest and yet have non-zero acceleration.Reason : The zero velocity of a particle at any instant does not necessarily imply zero acceleration at that instant.A. If both assertion and reason are trure and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanations of assertion.C. If assertion is true but reason is false.D. If both assertion and reason is false
Answer» Correct Answer - (a) A particle may be momentarily at rest and yet have non-zero acceleration. For example, a particle thrown up has zero velocity at its uppermost point but the acceleration at that instant continues to be the acceleration due to gravity.

Discussion

310.	A body is projected vertically up with velocity `98ms^(1)` after 2 s if the acceleration due to gravity of earth disappears, the velocity of the body at the end of next 3s isA. `49ms^(-1)`B. `49.6ms^(-1)`C. `78.4ms^(-1)`D. `94.7ms^(-1)`
Answer» Correct Answer - C `v=u-g t`. After g disappears body moves up with uiform velocity

Discussion

311.	A steam boat goes across a lake and comes back : (a) on a quiet day when the water is still and (b) on a rough day when there is a uniform current so as to help the journey onward and to impede the journey backward. If the speed of launch on both days same, in which case will it complete the journey in lesser time?
Answer» If the length of the lake is L and the velocity of boat is V, time taken in going and coming back on a quiet day: `t_(Q) = (L)/(V) + (L)/(V) = (2L)/(V)` ....(i) Now if `v` is the velocity of air-current, then in going across the lake `t_(1) = (L)/((V + v))` [as current helps the motion] and time taken in coming back `t_(2) = (L)/((V - v))` [as current opposes the motion] So, `t_(R) = t_(1) + t_(2) = (2LV)/((V^(2) - v^(2))) = (2L)/(V[1 - (v//V)^(2)])`....(ii) So, from Eqns. (i) and (ii), `(t_(R))/(t_(Q)) = (1)/([1 - (v//V)^(2)]) gt 1` [as `1 - (v//V)^(2) lt 1`] i.e., `t_(R) gt t_(Q)` i.e., time taken to complete the hourney on a quiet day is less than on a rough day.

Discussion

312.	A packet is released from a balloon which is moving upward when the balloon is at a height 200 m above ground. The packet reaches the ground in 8 sec. Speed of the balloon when the packet is released, is: (Take `g = 10 m//s^(2)`)A. 18 m/sB. 15 m/sC. 12 m/sD. 9 m/s
Answer» Correct Answer - B `-200 = u xx 8 - (1)/(2) xx 10 xx (8)^(2)`

Discussion

313.	Two plane, smooth surfaces are parallel to each other and are initiallt a distance of 2 metre apart. The two surfaces approach each other with a velocity of `1 cm//sec`. A particle starts with a velocity of 4 cm/sec from one surface and collides normally and elastically on the other surface from the time the two surfaces start moving. The collisions continues back and forth till the surfaces touch each other. The total distance covered by the particle is :A. 2 mB. 1 mC. 4 mD. 3 m
Answer» Correct Answer - C

Discussion

314.	A train is moving towards East with a speed 20 m/s. A person is running on the roof of the train with a speed 3 m/s against the motion of train. Velocity of the person as seen by an observer on ground will be :A. 23 m/s towards EastB. 17 m/s towards EastC. 23 m/s towards WestD. 17 m/s towards
Answer» Correct Answer - B Velocity of person as seen by observer `= 20 - 3 = 17 m//s`

Discussion

315.	The variation of quantity A with quantity B is plotted in the fig. Describes the motion of a particle in a straight line. (a) Quantity B may represent time. (b) Quantity A is velocity if motion is uniform. (c) Quantity A is displacement if motion is uniform (d) Quantity A is velocity if motion is uniformly accelerated.A. Quantive B may represent timeB. Quantity A is velocity if motion is uniformC. Quantity A is displacement if motion is uniformD. Quantify A is velocity if motion is uniformly acclerated
Answer» Correct Answer - A::C::D When we are calculating velocity of a displacement-time graph we have to take slope similarly we have to take slope of velocity-time graph to calculate accelration. When slope is constant motion wil be uniform . When we are representing motion by a graph it may be displacement -time velocity-time or aceelration-time hence. B may represent time. For uniform motion velocity-time graph should will be a stright line parallel to time axis. For uniform motion velocity is constant hence, slope will be positive. Hence quantity A is displacement. For uniformaly accelerated motion slope will be positive and A will represent velocity.

Discussion

316.	The variation of quantity A with quantity B is plotted in the fig. Describes the motion of a particle in a straight line. (a) Quantity B may represent time. (b) Quantity A is velocity if motion is uniform. (c) Quantity A is displacement if motion is uniform (d) Quantity A is velocity if motion is uniformly accelerated.A. Quantity B may represent timeB. Quantity A is velocity if motion is uniformC. Quantity A is displacement if motion is uniformD. Quantity A is velocity if motion is uniformly accelerated.
Answer» Correct Answer - A::C::D When we are calculating velocit of a displacement time graph we have to take slope similary we have to take slope of velocity-time graph to calculate acceleration when slope is constant motion will be uniform. When we are representing motion by a graph it may be displacement time, velocity-time or acceleration-time hence, B may represent time for uniform motion velocity-time graph should be a straight line parallel to time axis. For uniform motion velocity is constant hence, slope will be positive. Hence quantity A is displacement For uniformly accelerated motion slope will be positive and A represent velocity.

Discussion

317.	For a uniform motion.A. the velocity-time graph is a straight line parallel to time axisB. the position-time graph is a parabolaC. the acceleration-time graph is a straight line inclined with time axisD. the position-time graph is a straight line
Answer» Correct Answer - A

Discussion

318.	The variation of quantity A with quantity B. plotted in the fig. Describes the motion of a particle in a straight line. (a) Quantity B may represent time. (b) Quantity A is velocity if motion is uniform. (c) Quantity A is displacement if motion is uniform (d) Quantity A is velocity if motion is uniformly accelerated.A. a,c,dB. b,c,dC. a,bD. c,d
Answer» Correct Answer - A

Discussion

319.	The displacement of a particle as a function of time is shown in . It indicates .A. the particle starts with a certain velocity but the motion is retarded and finally the particle stopsB. the velocity of the particle is constant throughoutC. the acceleration of the particle is constantD. the particle starts with a constant velocity, the motion is accelerated and finally the particle moves with another constant velocity
Answer» Correct Answer - A `(ds)/(dt) = 0`, finally the particle comes at rest.

Discussion

320.	The displacement of a particle as a function of time is shown in . It indicates .A. The particle starts with a certain velocity but the motion is retarded and finally the particle stopsB. The velocity of the particle decreasesC. The acceleration of the particle is in opposite direction to the velocityD. The particle starts with a constant velocity the motion is accelerated and finally the particle moves with another constant velocity.
Answer» Correct Answer - A::B::C Initially at origin slope is not zero, so the particle has some initial velocity but with time we see that slope is decreasing and finally the slope becomes zero, so the particle stops finally.

Discussion

321.	Figure shows the displacement time graph of a particle moving on the X-axis A. The particle is continuously going in positive X direction.B. the particle is at restC. the velocity increases up to a time `t_(0)` and then becomes constant.D. the particle moves at constant velocity up to a time `t_(0)` and then stops.
Answer» Correct Answer - D

Discussion

322.	The Figure-1.109 shows the displacement-time graph ofa body subject only to the force of gravity. This graph indicates that: A. at A, acceleration = 0B. at A, velocity = maximumC. at A, displacement = 0D. the acceleration is constant at all the time
Answer» Correct Answer - D Since `s - t` graph is parabolic hence acceleration is constant

Discussion

323.	The motion of a body is given by the equation d`nu`/dt = 6 - 3`nu` where `nu` is the speed in `m s^(-1)` and t is time in s. The body is at rest at t = 0. The speed varies with time asA. `nu = (1 - e^(-3t))`B. `nu = 2(1 - e^(-3t))`C. `nu = 1 + e^(-2t)`D. `nu = 2(1 + e^(-2t))`
Answer» Correct Answer - B `(dv)/(dt) = 6 - 3v` or `(dt) = (dv)/(6 - 3v)` Integrating both sides, we get `t = 1/3 ln(6 - 3v) + C` where C is a constant of integration At t = 0, v = 0 `therefore` `C = 1/3 ln 6` `therefore = -1/3 ln(6 - (3v)/(6)` or `e^(-3t) = 1 - 1/(2 v)` or `v = 2(1 - e^(-3t))`

Discussion

324.	Figure shows the displacement time graph of a particle moving on the X-axis A. The particle is at rest.B. The particle is continuously going along x-direction.C. The velocity of the particle increases upto time `t_o` and then becomes constant.D. The particle moves at a constant velocity up to a time `t_o` and then stops.
Answer» Correct Answer - D The displacement-time graph is a straight line inclined to time axis upto time `t_o` indicates a uniform velocity. After time `t_o`, the displacement-time graph is a straight line parallel to time axis indicates particle at rest.

Discussion

325.	The position x of a particle with respect to time t along x-axis is given by `x=9t^(2)−t^(3)` where x is in metres and t is in seconds. What will be the position of this pariticle when it achieves maximum speed along the + x direction ?A. 54 mB. 81 mC. 24 mD. 32 m
Answer» Correct Answer - A Given, `x = 9t^(2) - t^(3)` Speed, `v = dx/dt = d/dt(9t^(2) - t^(3))= 18t - 3t^(2)` For maximum speed, `(dv)/(dt) = 0` `therefore` 0 = 18 - 6t or t = 3 s. `therefore x_"max" = 9 xx 3^(2) - 3^(3) = 81 - 27 = 54` m

Discussion

326.	The slope of the tangent at a point on the curve of concentration of a reactant as a function of time gives the instantaneous rate of reaction.A. accelerationB. velocityC. impusleD. momentum
Answer» Correct Answer - A Their slope of the tangent drawn on velocity-time graph at any instant of time is equal to the instantaneous acceleration.

Discussion

327.	Two trans one t ravelling at 54 kmph and the other at 72 kmph are headed towards one another along a straingt track. When they are `1//2` km apart, both drivers simultaneously seee the other train and apply their brakes. If each train is decelerated at the rate of `1 m//s^(2)`, will there be a collision ?
Answer» Velocity of the first train is 54 kmph`=15(m)/(s)` Distance travelled by the first train before coming to rest `s_(1)=(u^(2))/(2a)=(225)/(2)=112.5m` Velocity of the second train is `72 kmph = 20 m//s` Distance t ravelled by the second train before coming to rest `s_(2)=(u^(2))/(2a)=(400)/(2)=200m` Total distance travelled by the two trains before coming to rest `= s_(1)+s_(2)=112.5+200=312.5m` Because the initial distance of separation is 500 m which is greater than 312.5 m, there will be no collision between the two trains.

Discussion

328.	A bus accelerated from rest at a constant rate `alpha` for some time, after which it decelerates at a constant rate `beta` to come to rest. If the total time elapsed is t seconds. Then evaluate following parameters from the given graph (a) the maximum velocity achieved. (b) the total distance travelled graphically and (c) Average velocity.
Answer» (a). `alpha="slope of line" OA=(v_(max))/(t_(1))impliest_(1)=(v_(max))/(alpha)` `beta="slope of line" AB=(v_(max))/(t_(2))impliest_(2)=(v_(max))/(beta)` `t=t_(1)+t_(2)=(v_(max))/(alpha)+(v_(max))/(beta)=v_(max)((alpha+beta)/(alphabeta))` `v_(avg)=((alphabeta)/(alpha+beta))` (b). Displacement`=` area under the v-t graph `=(1)/(2)xx "base" xx "height" =(1)/(2)(t_(1)+t_(2))v_(max)=(1)/(2)tv_(max)` `=(1)/(2)((alphabeta)/(alpha+beta))t^(2)` `v_("avg")=("total displacement")/("total time")=(1)/(2)((alpha beta)/(alpha+beta))t=(v_(max))/(2)`

Discussion

329.	A body is projected vertically up with a valocity u. its velocity at ghalf of its maximum height and at `3//4th` of its maximum height are.
Answer» From `v^(2)-u^(2)=2aS`, here `a=-g,` when `S=(H)/(2)`, then `v^(2)-u^(2)=2(-g)(u^(2))/(4g)impliesv=(u)/(sqrt(2))` When `S=(3H)/(4)`, then `v^(2)-u^(2)=2(-g)(3u^(2))/(4(2g))impliesv=(u)/(2)`

Discussion

330.	A body thrown vertically upwards with an initial valocity `u` reaches maximum height in 6 s. The ratio of the distances traveled by the body in the first second the seventh second isA. `1:1`B. `11:1`C. `1:2`D. `1:11`
Answer» Correct Answer - B Velocity becomes zero after `6 s`. `0=u-gxx6impliesu=60 m//s` Distance in the first second `s_(1)=ut-1/2g t^(2)=60xx1-1/2xx10xx1^(2)=55 m` (i) After `6 s`, particle is moving in downward direction. (ii) Distance in the `7^(th)` second, `s_(2)=(1)/(2).10.1^(2)=5 m` `s_(1)/s_(2)=55/5=11/1`

Discussion

331.	A body is thrown vertically upward with velocity `u`. The distance traveled by it in the fifth and the sixth second are equal. The velocity `u` is given byA. `25 m//s`B. `50 m//s`C. `75 m//s`D. `100 m//s`
Answer» Correct Answer - B Since distance in the `5^(th)` and `6^(th)` second is equal, i.e., velocity after `5 s` becomes zero. `0=u-gxx5impliesu=50 m//s`

Discussion

332.	A body starts from rest. What is the retio of the distance traveled by the body during the `4^(th)` and `3^(rd)` secondsA. `7/5`B. `5/7`C. `7/3`D. `3/7`
Answer» Correct Answer - A

Discussion

333.	A ball is dropped vertically from `a` height `d` above the ground . It hits the ground and bounces up vertically to a height ` (d)//(2). Neglecting subsequent motion and air resistance , its velocity `v` varies with the height `h` above the ground asA. B. C. D.
Answer» Correct Answer - A For downward journey : `v^(2) = u^(2) + 2gh` Therefore, the velocity will increase in downward (negative) direction. The shape of graph will be parabolic. For upward journey : `v^(2) = u^(2) - 2gh` therefore, the velocity of the ball will become zero at the highest point. Again, the graph will be parabolic. option (a) is correct

Discussion

334.	A small block slides without friction down an iclined plane starting form rest. Let `S_(n)` be the distance traveled from time `t = n - 1` to `t = n`. Then `(S_(n))/(S_(n + 1))` is:A. `(2n - 1)/(2n)`B. `(2n + 1)/(2n - 1)`C. `(2n - 1)/(2n + 1)`D. `(2n)/(2n + 1)`
Answer» Correct Answer - C `S_(n) = (a)/(2) (2n - 1) , S_(n + 1) = (a)/(2) (2n + 1)` `(S_(n))/(S_(n + 1)) = (2n - 1)/(2n + 1)`

Discussion

335.	A stone is dropped from the top of a tall cliff and `n` seconds later another stone is thrown vertically downwards with a velocity `u`. Then the second stone overtakes the first, below the top of the cliff at a distance given byA. `g/2[(n(u - (gn)/2))/((u - gn))]^(2)`B. `g/2[(n(u/2-gn))/(u-gn)]^(2)`C. `g/2[(n(u/2-gn))/(u/2-gn)]^(2)`D. `g/5[(u-gn)/(u/2-gn)]^(2)`
Answer» Correct Answer - A Let the two stones meet at time t. For the first stone, `S=1/2"gt"^(2) (therefore u=0)`……………(i) For the second stone, `s_(2)=u(t-n)+1/2g(t-n)^(2)`…………….(ii) Displacement is same `therefore S_(1)=S_(2)` `1/2"gt"^(2)=u(t-n)+1/2(g(tn)^(2)` (Using (i) and (ii)) `1/2"gt"^(2)=ut-nu+1/2"gt"^(2)+1/2"gn"^(2)-"gtn"` `ut-"gtn"="nu"-1/2"gn"^(2)` `t=("nu"-1.2"gn"^(2))/(u-"gn") = (n(u-g)/(2n))/(u-gn)` or `S_(1) = g/2[(n(u-(gn)/(2)))/(u-gn)]^(2)` from (i)

Discussion

336.	An object is thrown vertically upward with a speed `u_(1)` and it travels 8 m in the last secon of its upward motion. If the object is thrown upward with a speed `u_(2)` which is twice of `u_(1)`, the distance now travelled by the object during the last second of its upward journey will be:A. 32 mB. 16 mC. 12 mD. 8 m
Answer» Correct Answer - D The distance travelled in last second is always equal because it is equal to `s = (1)/(2) xx g (t)^(2)`

Discussion

337.	A body is thrown up with a velocity `100ms^(-1)` it travels 5 m in the last swecond of upward journey if the same body thrown up with velocity `200ms^(-1)`, how much distance (in metre) will it travel in the last second of its upward journey `(g=10ms^(-2))`
Answer» Correct Answer - 5 `s=u+(a)/(2)(2n-1)` `u=100ms^(-1),a=10ms^(-2)` and `s=5m` `5=10-5(2n-1)` gives `n=10s` Body thrown up with velocity `200ms^(-1)` will take 20 s to reach the highest point. Distance Travelled in `20^(th)` second is `200-5(20xx2-1)=5m` In the last second of upward journey, the bodies will travel same distance.

Discussion

338.	The distance travelled by a body during last second of its upward journey is d, when the body is projected with certain velocity vertically up. If the velocity of projection is doubled the distance travelled by the body during the last second of its upward journey isA. `2d`B. `4d`C. `d//2`D. `d`
Answer» Correct Answer - D Distance covered in the last second of upward journey`=` distance covered in 1 st second of downward journey. This is independent of velocity of projection.

Discussion

339.	An express train moving at `30 m//s` reduces its speed to `10 m//s` in a distance of `240m`. If the breaking force is increased by `12.5%` in the begininning find the distance that it travels before coming to restA. 270 mB. 240 mC. 210 mD. 195 m
Answer» Correct Answer - A `V^(2)-u^(2)=2as`

Discussion

340.	Two particles start simultaneously from the same point and move along two straight lines. One with uniform velocity v and other with a uniform acceleration a. if `alpha ` is the angle between the lines of motion of two particles then the least value of relative velocity will be at time given byA. `(v)/(a) sin alpha`B. `(v)/(a) cos alpha`C. `(v)/(a) tan alpha`D. `(v)/(a) cot alpha`
Answer» Correct Answer - B At any time velocity of first car is v and that of second car is `v = v + at = 0 + at` `v_("rel") = sqrt(v^(2) + (at)^(2) - 2v at cos alpha)` `v_("rel")` is minimum if `(d)/(dt) (v_(r)^(2)) = 0, t = (v cos alpha)/(a)`

Discussion

341.	A lift is coming from `8th` floor and is just about to reach `4th` floor. Taking ground floor as origin and positive direction upwards for all quantities, which one of the following is correct ?A. `xlt0,vlt0,agt0`B. `xgt0,vlt0,alt0`C. `xgt0,vlt0,agt0`D. `xgt0,vgt0,agt0`
Answer» Correct Answer - A As the lift is coming iin downward direction displacement will be negative we have to see whether the moton is accelerating or retarding we know that due to downwards motion displacement will be negative when the lift reaches 4th floor is about to stop hence, motion is retarding in nature hence, `xlt0,agt0`. As displacement is in negative direction, velocity will also be negative i.e., `Vlt0` This can be shown on the adjacent graph.

Discussion

342.	A lift is coming from `8th` floor and is just about to reach `4th` floor. Taking ground floor as origin and positive direction upwards for all quantities, which one of the following is correct ?A. x gt 0, v lt 0, a gt 0B. x gt 0, v lt 0, a lt 0C. x gt 0, v lt 0 , a gt 0D. x gt 0, v gt 0 a lt 0
Answer» Correct Answer - (a) As lift is coming from `8^(th) to 4^(th)` floor, it is retarding, i.e., a is acting upwards, so, a gt 0 and the value of x becomes less hence negative, i.e., x lt 0, velocity is downwards(i.e., negative) so, v lt 0.

Discussion

343.	A point moves linearly with deceleration which is given by `dv//dt=-alphasqrt(v)`, where alpha is a positive constant. At the start `v=v_(0)`. The distance traveled by particle before it stops will be
Answer» `(dv)/(dt)=-alphav^(1//2)` `int_(v_(0))^(v)v^(-1//2) dv=-alphaint_(0)^(t) dt` `(\|v^(-1/2+1)\|_(v_(0))^(v))/((-1/2+1))=-alpha\|t\|_(0)^(t)` `\|v^(1/2)\|_(v_(0))^(v)=alpha/2t` `sqrt(v)-sqrt(v_(0))=-alpha/2timpliesv=[sqrt(v_(0))-(alphat)/2]^(2)` The particle will stop when `v=0` `t=(2sqrt(v_(0)))/alpha` To calculate the displacement by velocity is lengthy. `(dv)/(dt)=v(dv)/(ds)=-alpha v^(1//2)` `int_(v_(0))^(v) v^(1//2) dv=-alphaint_(0)^(s) ds` `\| v^(3//2)\|_(v_(0))^(v)/(3/2)=-alphas` `v^(3//2)-v_(0)^(3//2)=-3/2alphas` When particle stops, `v=0` `impliess=(2v_(0)^(3//2))/(3alpha)`

Discussion

344.	The velocity of a particle moving along the xaxis is given by `v=v_(0)+lambdax`, where `v_(0)` and lambda are constant. Find the velocity in terms of t.
Answer» `v=v_(0)+lambdax` `(dv)/(dt)=0+lambda(dx)/(dt)=lambdav` `a=lambdav` At `x=0, v=v_(0)` `a=lambdav` `(dv)/(dt)=lambdav` `int_(v_(0))^(v)(dv)/v=lambda int_(0)^(t) dt` `\|log_(e)v\|_(v_(0))^(v)=lambda\|t\|_(0)^(t)` `log_(e)v-log_(e)v_(0)=lambdat` `log_(e)(v/v_(0))=lambdat` `v/v_(0)=e^(lambdat)` `v=v_(0)e^(lambdat)`

Discussion

345.	A particle is moving such that `s=t^(3)-6t^(2)+18t+9`, where s is in meters and t is in meters and t is in seconds. Find the minimum velocity attained by the particle.
Answer» `s=t^(3)-6t^(2)+18t+9` `v=(ds)/(dt)=3t^(2)-12t+18` For v to be minimum or maximum, `a=(ds)/(dt)=6t-12=0impliest=2 s` At `t=2 s`, `(d^(2)v)/(dt^(2))=6gt0`, i.e. v is minimum at `t=2 s` `v_(min)=3(2)^(2)-12(2)+18=6 m//s`

Discussion

346.	The position of particle moving along the x-axis veries with time t as `x=6t-t^(2)+4`. Find the time-interval during which the particle is moving along the positive x-direction.
Answer» `x=6t-t^(2)+4` `v=(dx)/(dt)=6-2t=2(3-t)` At `tlt3 s, vgt0`, the particle is moving along the positive x-direction. At `tgt 3 s, vlt0`, the particle is moving along the negative x-direction. At `t=3 s, v=0`. For time-interval `t=0` to `t=3 s`, the particle is moving along the positive x-direction.

Discussion

Explore topic-wise InterviewSolutions in .

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