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1.

The position of a particle moving on X-axis is given by `x =At^(2) + Bt + C` The numerical values of A, B and C are 7, -2 and 5 respectively and SI units are used. Find (a) The velocity of the particle at t= 5 (b) The acceleration of the particle at t =5 (c ) The average velocity during the interval t = 0 to t = 5 (d) The average acceleration during the interval t = 0 to t = 5

Answer» `x = 7t^(2) - 2t + 5`
(a) `v = (dx)/(dt) = 14t - 2`
at `t=5, v = 14 xx 5-2 = 68 m//s`
(b) `a= (dv)/(dt) = 14 m//s^(2)`
(c ) Average velocity = `("displacement")/("Time")= (x_5- x_0)/(5-0)`
`x_5 = 7(5)^(2) - 2(5) + 5 = 170 `m
`x_(0) = 7(0)^(2)- 2(0) + 5 = 5`m
`v_("avg") = (170-5)/( 5) = 33 m//s`
(d) Average acceleration
`" " = ("Change in velocity")/("Time interval") = (v_5 - v_0)/( 5-0)`
`v_5 = 14 xx 5- 2 = 68 m//s`
`v_0 = 14 xx 0 -2 = -2m//s`
`a_("avg") = (68-(-2))/(5-0) = 14 m//s^(2)`
Discussion
2.

A particle goes from point A to point B, moving in a semicircle of radius 1 m in 1 second. Find the magnitude of its average velocity.

Answer» Average velocity
`" "= ("Net displacement")/("total time")= (AO + OB)/(time)`
`" "=(1+1)/(1) = 2m//s`
Discussion
3.

A particle is dropped from the top of a tower. During its motion it covers `(9)/(25)` part of height of tower in the last 1 second Then find the height of tower.

Answer» Let it taken n seconds to fall from tower then total distance `s = (1)/(2) g(n)^(2)`
distance travelled in `n^(th)` second
`s_(n^(th))= O + (g)/(2) (2n-1) rArr s_(n^(th)) = (9)/(25) s `
`rArr (g)/(2) (2n -1) = (9)/(25) (g)/(2)(n^(2)) rArr 9n^(2) = 50 n - 25 `
`rArr n = (50 + sqrt(50^(2) - 4 xx 9xx (+ 25)))/( 2xx 9) = 5 sec`.
`H = (1)/(2) g(5)^(2) = 125 `m
Discussion
4.

The distance travelled by a particle in time t is given by `s = (2.5 t^(2))` m. Find (a) the average speed of the partion during time 0 to 0.5 s and (b) the instantaneous speed at t = 5.0s.

Answer» (a) The distance travelled during time 0 to 5.0 s is,
`S = (2.5)(5.0)^(2) = 62.5` m
The average speed during this time is
`v_("avg")= ("distance travelled")/("time taken") = (62.5)/(5) = 12.5 m//s`
(b) Given `S = (2.5)t^(2)`
Intantaneous speed = `(dS)/(dt) = (d)/(dt) (2.5t^(2)) = (2.5)(2t) = 5t`
`therefore ` At `t = 5.0s` the speed is
`[v]_(t=5S)= (5.0) xx (5.0) = 25 m//s`
Discussion
5.

A body is dropped from a height h above the ground. Find the ratio of distances fallen in first one second, firt two seconds, first three seconds, also find the ratio of distance fallen in `1^(st)` second, in `2^(nd)` second, in `3^(rd)` second etc.

Answer» From second equation of motion, i.e h = `(1)/(2) "gt"^(2)" "(h= ut + (1)/(2)"gt"^(2) and u=0)`
`h_1 : h_2 : h_3 ...... = (1)/(2) g(1)^(2) : (1)/(2)g(3)^(2) = 1^(2) : 2^(2) : 3^(2) .......... = 1 : 4 : 9 : ....... . `
Now from th expression of distance travelled in `n^(th)` second `S_n = u + (1)/(2)a (2n-1)`
here `u =0, a= g" "` So `" "S_n= (1)/(2)g(2n-1)` therefore
`S_1 : S_2 : S_3 ......... = (1)/(2) g ( 2xx1- 1) : (1)/(2)g (2xx 2 -1) : (1)/(2)g( 2xx 3 -1)= 1 : 3: 5 .............. `
Discussion
6.

The velocity of a particle is given by `v=(2t^(2)-4t+3)m//s` where t is time in seconds. Find its acceleration at t=2 second.

Answer» Acceleration (a) `= (dv)/(dt) = (d)/(dt) (2 t^(2) - 4t+ 3)= 4t -4`
Therefore acceleration at `t = 2s` is equal to, `a = ( 4xx2) - 4 = 4 m//s^(2)`
Discussion
7.

A particle moves on straight line according to the velocity-time graph shown in fig. Calculate - (i) Total distance covered (ii) Average speed (iii) In which part of the graph the acceleration is maximum and also find its value. (iv) Retardation

Answer» (i) Total distance covered = Area under v - t curve
` " "= (1)/(2) (2xx2) + 2(4-2) + (1)/(2) (10+2) xx 1`
`(1)/(2) xx 5 xx 10 = 37` m
(ii) Average speed = `("Total distance")/("Total time") = (37)/(10) = 3.7 m//s`
(iii) Acceleration = Slope of v-t curve
So maximum acceleration will be in the part where the slope will be maximum i.e. BC
`a_(max) = a_(BC) = (10-2)/(1) = 8m//s^(2)`
(iv) Retardation = Slope of CD (As it is negative)
`" " = |(0-10)/(5)| = |-2| = 2m//s^(2)`
Discussion
8.

Straight distance between a hotel and a railway stations is 10 km, but circular route is followed by a taxi coven 23 km in 28 minute. What is average speed and magnitude of average velocity ? Are they equal ?

Answer» (a) Average speed= `("Total path length")/("Total time")`
`=(28)/((28//60)) = 49.3 kmh^(-1)`
(b) Average velocity= `("Displacement")/("Time")`
`= ("Shortest distance")/("Time") = (10)/(28// 60) = 21. 4 km//h`
(c ) Both are Not equal
Discussion
9.

For a particle moving with constant acceleration, prove that the displacement in the `n^(th)` second is given by `s_(n^(th)) = u + (a)/(2)(2n-1)`

Answer» From `s= ut + (1)/(2) at^(2)`
`" "s_n =un + (1)/(2) an^(2) ………. (1)`
`" "s_(n-1) = u(n-1) + (1)/(2) a (n-1)^(2)………. (2)`
By equation (1) & (2)
`s_n- s_(n-1)= s_(n^(th))= u + (a)/(2)( 2n-1)`
Discussion
10.

A cricketer can throw a ball to a maximum horizontal distance of 100 m. How high above the ground can the cricketer throw the ball, with the same speed ?

Answer» Let u be the velocity of projection of the ball. The ball will cover maximum horizontaal distance when angle of projection with horizontal , `theta = 45^(@)`. Then
`R_(max) = (u^(2))/(g) = 100 ` m
If ball is projected vertically upwards `(theta =90^(@) ` from ground `)` then H attains maximum value,
`H_(max) =(u^(2))/(2g) = (R_(max))/(2)`
`therefore ` the height to which cricketer can throw the ball is `= (R_(max))/(2) = (100)/(2) = 50 `m
Discussion
11.

A football player kicks a ball at ball at an angle of `30 ^(0)` with the horizontal with an initial speed of `20 m//s`. Assuming that the ball travels in a vertical plance, calculate (a) the time at which the ball reaches the highest point (b) maximum height reached (c ) the horizrontal range of the ball (d) the time for wich the ball is in air. `g =10 m//s^(2)`.

Answer» (a) Time taken by the ball to reach the highest point
`t = (T)/(2) = (u sin theta )/(g) = (20)/(10) xx sin 30^(@) = 2 xx (1)/(2) = 1`s
(b) The maximum height attained
`= (u^(2) sin^(2) theta)/( 2g) = ((20)^(2) xx sin^(2) 30^(@))/( 2xx 10) = 5` m
(c) The horizontal range
`= (u^(2) sin 2 theta )/(g) = ((20)^(2) xx sin( 2xx 30^(@)))/(10) = 34. 64` m
(d) The time of flight
`(2u sin theta)/(g) = ( 2xx 20 xx sin 30^(@))/(10) = 2s `
Discussion
12.

Two cars start off a race with velocities 2m`//`s and 4m`//`s travel in straight line with uniform acceleration `2m//s^(2) and 1m//s^(2)` respectively. What is the length of the path if they reach the final point at the same time ?

Answer» Let both particles reach at same position in same time t then from s = `ut + (1)/(2) at ^(2)`
For `1^(st)` particle : `s = 4(t)+ (1)/(2)(1) t^(2) = 4t + (t^(2))/(2)`, For `2^(nd)` particle : `s = 2(t) + (1)/(2) (2)t^(2) = 2t + t^(2)`
Equation above equation we get `4t+ (t^(2))/(2) = 2t + t^(2) rArr t = 4s`
Subsituting value of t in above equation s = `4(4) + (1)/(2) (1)(4)^(2)= 16 +8 = 24`m
Discussion
13.

Air distance between Kota to Jaipur is 260 km and road distance is 320 km. A deluxe bus which moves from Jaipur to Kota takes 8 h while and aeroplane reaches in just 15 min. Find (i) average speed of bus in km/h (ii) average velocity of bus in km/h (iii) average speed of aeroplane in km/h (iv) average velocity of aeroplane in km/h

Answer» Air distance (shortest distance) = displacement = 260 km
Road distance = 320 km
(i) Average speed of bus = `("Total distance")/("total time")`
`" " = (320)/(8)`
`" " = 40 km//h`
(ii) Average velocity of bus
`" " = ("Total displacement")/("Time")= (260)/(8)`
`" " = 32.5 km//h`
(iii) Average speed of plane = `(260)/((1)/(4)) = 1040 km//h`
(iv) Average velocity of aeroplane = `(260)/(1//4)`
`" " = 1040 km//h`
Discussion
14.

A particle is projected from the ground at an angle such that it just clears the top of a polar after `t_1` time its path. It takes further `t_2` time to reach the ground. What is the height of the pole ?

Answer» Height of the pole is equal to the vertical displacement of the particle at time `t_1`
Vertical displacement `y= ut_1 + (1)/(2) a t_(1)^(2) = ut_1 - (1)/(2) "gt"_(1)^(2)………….` (i)
and total flight time `t_1 + t_2 = ( 2u_2)/(g) rArr u_1 = (g)/(2) (t_1 +t_2)`
put value `u_2` in equation (i) `y = (g)/(2)(t_1 + t_2) t_1 - (1)/(2) g(t_1)^(2) = (1)/(2)"gt"_1 t_2`. so height of the pole `= (1)/(2) "gt"_1t_2`.
Discussion
15.

A rocket is fired vertically up from the ground with a resultant vertical acceleration of `10 m//s^2.` The fuel is finished in 1 min and it continues to move up. (a) What is the maximum height reached? (b) Afte2r how much time from then will the maximum height be reached?(Take `g= 10 m//s^2`)

Answer» (a) The distance travelled by the rocket during burning interval (1 minute = 60 s) in which resultant acceleration is vertically upwards and `10 m//s^(2) ` will be `h_1 = 0 xx 60 + (1//2) xx 10 xx 60^(2) = 18000 `m = 18 km and velocity acquired by it will be `v= 0 + 10 xx 60 = 600 m//s`
Now after 1 minute the rocket moves vertically up with initial velocity of 600 m/s and acceleration due to gravity opposes its motion. So, it will go to a height `h_2` from this point, till its velocity becomes zero such that `0 = (600)^(2) - 2gh_2 or h_2 = 18000` m = 18 km [ g= 10 `m//s^(2)`]
So the maximum height reached by the rocket from the ground, `H = h_1 + h_2 = 18 + 18 = 36` km
(b) As after burning of fuel the initial velocity 600 m/s and gravity opposes the motion of rocket , so from `1^(st)`
`0 = 600 - "gt" rArr t = 60 s`
Discussion