Saved Bookmarks
This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A long capillary tube of radius r dipped in water such that sqrt(3/4)of the height it can rise in thecapillary is available above free level of water the angle of contact at the top level in the tube is pi/k What is value of k. |
|
Answer» |
|
| 2. |
Is it possible that at any moment velocity of any object moving in straight line is zero and acceleration is non-zero ? Give example. |
| Answer» SOLUTION :Yes, when an object is THROWN upwards and it REACHES at MAXIMUM height, its velocity becomes zero and its acceleration remains CONSTANT. | |
| 3. |
How is the period of a pendulum affected when its point of suspension is moved vertically downward with a acceleration a < g. |
| Answer» Solution :When the POINT of suspension is moved VERTICALLY downward, then effective acceleration is (G — a). So T will INCREASE | |
| 4. |
A boat with bricks is floating on the water. When the bricks are thrown into water, the level of water in the pond: |
|
Answer» rises |
|
| 5. |
An alternative name of 'mean calorie' is |
|
Answer» `0^(@)-100^(@)"C CAL"` |
|
| 6. |
What is the work done in increasing the angular frequency of a circular ring of mass 2kg and radius 25 cm from 10 rpm to 20 rpm about ist axis? |
|
Answer» SOLUTION :WORK DONE = Increase in rotational kinetic energy `=(1)/(2)l(omega_(f)^(2)-omega_(i)^(2))=(1)/(2)MR^(2)(omega_(f)^(2)-omega_(i)^(2))` `=(1)/(2)xx2xx(0.25)^(2)(((2pi)/(3))^(2)-((pi)/(3))^(2))=0.2058J` |
|
| 7. |
Inelastic collision is due to |
|
Answer» CONSERVATIVE FORCE |
|
| 8. |
A mass 'M' is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period 'T, if the mass is increased by 'm' time period becomes (5T)/(3) then the ratio (m//M) is |
|
Answer» `(3)/(2)` |
|
| 9. |
What is angle between instantaneous velocity and frictional force for object moving on rough surface ? |
| Answer» SOLUTION :`180^(@)`becausefrictional FORCEIS alwaysinoppositedirectionof MOTION | |
| 10. |
A mass of 0.1 kg is rotated in a vertical circle using a string of legth 20 cm. When the string makes an angle 30^(@) with the vertical, the speed of the mass is 1.5ms^(-1). The tangential acceleration of the mass at that instant is |
|
Answer» `4.9 MS^(-2)` |
|
| 11. |
Figure 14.26 (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure 14.26 (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. 14.26(b) is stretched by the same force F. If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case ? |
| Answer» SOLUTION :`T=2pisqrt(m/k)` for (a) and `2pisqrt(m/(2K))` for (B) | |
| 12. |
Figure 14.26 (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure 14.26 (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. 14.26(b) is stretched by the same force F. What is the maximum extension of the spring in the two cases ? |
| Answer» SOLUTION :F/k for both (a) and (B). | |
| 13. |
Which of the following energy-time graph represents damped harmonic oscillator ? |
|
Answer»
|
|
| 14. |
In an experiment to estimate the size of a molecue of oleic acid 1mL of oleic acid is dissolved in 19 mL of alochol. Then 1 mL of this solution is dilluted to 20 mL by adding alcohol. Now 1 drop of this diluted solution is placed on water in a shallow trough. The solution spreads over the surface of water forming one molecule thick layer. Now, lycopodium powder is sprinkled evenly over the film and its diameter is measured. Knowing the volume of the drop and area fo the film we can calculate the thickness of the film which will give us the size of oleic acid molecule. Read the passage cearfully and answer the following questions : (a) why do we dissolve oleic acid in alcohol ? (b) What is the role of lycompodium powder ? (c ) What would be the volume of oleic acid in each mL of solution prepared ? (d) How will you calculate the volume of n drops of this solution of oleic acid ? (e ) What will be the volume of oleic acid in one drop of this solution ? |
|
Answer» Solution :(a) Oleic acid is DISSOLVED in alcohol, because it does no dissolve in water. (b) When LYCOPODIUM POWDER is sprinkled evenly, it spreads over the entire surface of water. When a drop of perpared solution is dropped on water, oleic acid does not dissolve in water. Instead, it spreads on the water surface pushing the lycopodium powder away to clear a circular area where the drop falls. We can therefore, measure the area over which oleic acid spreads. (c ) Volume of oleic acid in each mL of solution prepared ` (1)/(20)mLxx(1)/(20) = (1)/(400)mL` (d) Volume of n drops of this solution of oleic acid can be calculated by using a burette and measuring CYLINDER. (e ) If 1 mL of solution contains n drops, then volume of oleic acid in one drop ` = (1)/((400n)) mL` |
|
| 15. |
A 2 mu F condenser is charged upto 200 volt and then battery is removed. On combining this with another uncharged condenser in parallel, the potential difference is found to be 40 volt. Find the capacity of second condenser (in mu F) |
|
Answer» |
|
| 16. |
Find the number of possible natural oscillations of air column in a pipe frequencies of which lie belowv_(0) = 1250 Hz. The length of the pipe is l = 85 cm. The velocity of sound is v = 340 m//s. Consider two cases i. the pipe is closed from the end , ii. the pipe is open from both ends. |
|
Answer» Solution :First Method i. Pipe is closed from one end : An air column in a pipe closed from one end oscillates only odd harmonics[ 1st harmonic (fundamental mode ), 3rd harmonic ( 1st overtone ), 5th harmonic ( 2nd overtone ) , 7th harmonic ( 3rd overtone ) etc.] Fundamental frequency ` = (v)/( 4 l) = ( 340 )/( 4 xx (85)/(100)) = 100 Hz` Other modes of oscillation are : `3rd "harmonic frequency" = 3 xx 100 = 300 Hz` ` 5th "harmonic frequency"= 5 xx 100 = 500 Hz` `7th "harmonic frequency" = 7 xx 100 = 700 Hz` ` 9th "harmonic frequency" = 9 xx 100 = 900 Hz` `11th "harmonic frequency" = 11 xx 100 = 1100 Hz` `13th "harmonic frequency" = 13 xx 100 = 1300 Hz` Only those natural oscillations are to be counted frequencies of which lie below ` v_(0) = 1250 Hz`, the harmonics till 11th harmonic are to be counted. Since number of possible natural oscillations ` = 1 ( 1st "harmonic") + 1 ( 3rd "harmonic") + 1 ( 5th "harmonic" ) + 1 ( 7th "harmonic" ) + 1( 9th "harmonic" ) + 1 ( 11th "harmonic" ) = 6` second Method All the frequencies possible are integral MULTIPLES of fundamental frequency which is `100 Hz`. Using the fact that integer which is multiplied by fundamental frequency is the number of harmonic itself you get , highest harmonic predicted `= [ 12.50 // 100 ] ` where `[ x]` represents greatest integer less than or equal to ` x = [ 12.5 ] = 12`. Now for closed pipe , only odd harmonics are possible and the highest harmonic possible = 11 th . The possible harmonics are `1 , 3 , 5 , 7 , 9 , 11` which are six in number. ii. Pipe opened from the both ends : Fundamental frequency ` = (V)/( 2 l) = ( 340)/( 2 xx 85) xx 100 = 200 Hz` Frequency of the other natural modes of oscillational are : ` 2nd "harmonic frequency" =2 xx 200 = 400 Hz` `3rd "harmonic frequency" = 3 xx 200 = 600 Hz` ` 4TH "harmonic frequency" = 4 xx 200 = 800 Hz` ` 5th "harmonic frequency"= 5 xx 200 = 1000 Hz` `6th "harmonic frequency" = 6 xx 200 = 1200 Hz` ` 7th "harmonic frequency" = 7 xx 200 = 1400 Hz` You have to count only those harmonics whose frequencies are below `1250 Hz`. All the harmonics till 6 th harmonic are possible , and obviously they are six in number. Third Method Fundamental frequency ` = 200 Hz` Frequencies possible ` = n xx "fundamental frequency"` ` = n xx 200 ``[ n is 1 , 2, ...]` Maximum value of ` n = [ 12.50 // 200] = 6 ([x] "represents greater than or equal to" x)` Now ` n` is also equal to the number of harmonics for which frequency is being calculated , highest harmonic possible `= 6 th`. As all harmonics are possible in case of open tube , harmonics possible are `1st, 2nd , 3rd , 4th , 5th and 6th`. Number of harmonics possible in this case ` = 6`. |
|
| 17. |
A block of mass 10 kg , moving in x direction with a constant speed of 10 ms^(-1), is subjected to a retarding force F = 0.1 x J//m during its travel from x = 20m to 30 m. Its final KE will be |
|
Answer» Solution : Here, `10kg, v_(i)=10ms^(-1)` Initial kinetic energy of the block is `K_(i)=(1)/(2)mv_(i)^(2)=(1)/(2)xx(10kg)xx(10ms^(-1))^(@)=500J` Work done by retarding force `W=int_(x_(1))^(x_(2))F_(r)dx=int_(20)^(30)-0.1xdx=-0.1[(x^(2))/(2)]_(20)^(30)` `=-0.1[(800-400)/(2)]=-25J` According to work - energy theorem, `W=K_(f)-K_(i)` `K_(f)=W+K_(i)=-25J+500J=475J` |
|
| 18. |
The x and y displacementof a particle in the x-y plane at any instant are given by x=alphaT^(@) and y=2alphaT where a is a constant. The velocity of the particle at any instant is given by: |
|
Answer» `4asqrt(T^(2)+4)` |
|
| 19. |
A chain of length 1 and mass m lies on the surface of a smooth sphere of radius Rgt1 with one end tied to the top of the space. Suppose the chain is released and slides down the sphere. Find the kinetic energy of the chain, when it has slid through an angle theta. |
|
Answer» `(mR^(2)G)/LSIN (L/R)` |
|
| 20. |
A chain of length 1 and mass m lies on the surface of a smooth sphere of radius Rgt1 with one end tied to the top of the space. Find the gravitational energy of the chain with reference level at the centre of the sphere |
|
Answer» `(MR^(2)g)/LSIN(l/R)` |
|
| 21. |
For a surface molecule |
|
Answer» the net FORCE on it is NON zero |
|
| 22. |
Prove that gravitational field intensity at any point in equal to acceleration experienced at that point. |
|
Answer» Solution :The gravitational FIELD intensity `vec(E)_(1)` (here after called as gravitational field) at a point which is at a distance r from `m_(1)` is defined as the gravitational force experienced by unit MASS placed at that point. It is GIVEN by the ratio `(vec(F)_21)/(m_2)` (where `m_(2)` is the mass of the object on which `vec(F)_(21)` acts) `vec(F)_(21)=-(Gm_(1)m_(2))/(r^2)hat(r)""...(1)` Using `vec(E)_(1)=(vec(F)_(21))/(m_2)` (1) we get, `vec(E)_(1)=-(Gm_1)/(r^2)hat(r)""...(2)` `vec(E)_(1)` is a vector quantity that points towards the mass `m_(1)` and is independent of mass `m_(2)`. The value of `m_(2)` is taken to be of unit magnitude. The unit `hat(r)` is along the line between `m_(1)` and the point in question. The field `vec(E)_(1)` is due to the mass `m_(1)`. In general, the gravitational field intensity due to a mass M ATA distance r is given by `vec(E)=-(GM)/(r^2)hat(r)""...(3)` In the region of this gravitational field, a mass 'm' is placed at a point P, when mass 'm' interacts with the field `vec(E)` and experiences an attractive force due to M. The gravitational force experienced by 'm' due to 'M' is given by  `vecF_(m) = m vecE` Now we can equate this with Newton's second law `vec(F)=mvec(a)` `mvec(a)=mvec(E)` `vec(a)=vec(E)` |
|
| 23. |
If pressure P, velocity V and time T are taken as fundamental physical quantites, the dimensional formulae of the force is |
|
Answer» <P>`PV^(2)T^(2)` |
|
| 24. |
If eta =A/B log(Bx+C)is dimensionally true, then (here h is the coefficient of viscosity and x is the distance) |
|
Answer» C is dimensionless constant |
|
| 25. |
What is defined by dividing stress by strain within elastic limit? |
|
Answer» |
|
| 26. |
In reversible process the entropy of the universe |
|
Answer» increases |
|
| 27. |
Inside a compartment of a train running with a uniform velocity , a boy throws a ball. Does the kinetic energy of the ball depend on the velocity of the train? |
| Answer» SOLUTION :Here, the train RUNNING with a uniform velocity . So, if we considerthe train as the frame of reference the velocity and the kinetic energy of the ball becomes ZERO and does not depend on the velocity of the train. When the kinetic energy is calculated, takingthe earth.s SURFACE as the INERTIAL frame of reference, it will depend on the velocity of the train. | |
| 28. |
A freely body acquires a velocity of g^(x)h^(y) after falling through a height h. Using dimensions find the values fo x and y. |
|
Answer» Solution :`v=g^(X)h^(y)` Taking dimensions on both SIDES `LT^(-1)=(LT^(-2))^(x)(L)^(y)=L^(x)T^(-2x)L^(y)=L^(x+y)T^(-2x)` EQUATING the dimensions of T and L on both sides, `-2x=-1,x=1//2, 1=x+y,y=1-1//2=1//2,x=1/2,y=1/2` SUBSTITUTING in eq (i) `v=g^(1//2)R^(1//2),v=sqrt(gh)` |
|
| 29. |
The orbital radius of mercury around the sun is 0.38 A U. Determine the angle of maximum elogation for mercury and its distance from earth, where elognation is maximum, |
|
Answer» The angle of maximum elongation `"in"`is given by `SIN "in" = (r_(ps))/(r_(es)) = (0.38)/(1)` Now,`r_(pe) =r_(es)xxcos "in"` `1.496xx10^(11) cos (22.3^@)` `=1.496xx10^(11)xx0.9252 = 1.384xx10^(11)m` `=1.384xx10^8km` |
|
| 30. |
A point source emits sound equally in all direction is a non-absorbing medium. Two points p and Q are at distances of 2m and 3m, respectively, from the source. The ratio of the intensitiesof the waves at P and Q is |
|
Answer» `3 :2` |
|
| 31. |
Two bodies are thrown from the same point with the same velocity of 50 ms^(-1). If their angles of projection are complimentary angles and the difference of maximum heights is 30m, their maximum heights (g=10ms^(-2)) |
|
Answer» 50 m and 80 m |
|
| 32. |
Imagine a hypothetical planet "flat earth" of thickness .t. having same gravity on its surface as "our Earth". Assume "Flat earth" to be inifinite plate of same uniform densitty as "our Earth". Which is an uniform solid sphere of Radius R. IF t = (4R)/(n), find the value of n ? |
|
Answer» |
|
| 33. |
Three concetric sphereical metal shells A,B,C or radii a,b,c (c gt b gt a) have surface charge densities +sigma , - sigma and +sigma respectively. (a) Findthe potentialsof thethree shells? (b) If theshells A and Care at same potential , find therelation between a,b and c. |
|
Answer» Solution :(a) Chargeson the three SHELLS are `q_(A) + + sigma 4pia^(2)` `q_(B) = - sigma 4 PI b^(2)` `q_(B)= -sigma 4 oi b^(2)` `q_(C) = + sigma 4 pi c^(2)` `V_(A) = (1)/(4 pi in_(0)) {(q_(A))/(a) + (q_(B))/(b) + (q_(c))/(c)} = (sigma)/(in_(0)) (a+b+c)` `V_(B) = (1)/(4 pi in_(0)) { (q_(A))/(b) + (q_(B))/(b) + (q_(c))/(c)} = (sigma)/(in_(0)) ((a^(2)-b^(2))/(b) + c)` `V_(c)= (1)/(4PI in_(0)) {(q_(A))/(c) + (q_(B))/(c) + (q_(C))/(c)} = (sigma)/(in_(0))((a^(2) - b^(2) + c^(2))/(c))` (b) If `V_(A) = V_(C)` , on substitution . we geta+ b = c |
|
| 34. |
If the error in the measurement of radius of a sphere is 3% then the error in the measurement of volume of the sphere will be : |
|
Answer» Solution :`V=(4)/(3)pir^(3)` `DeltaV=3Deltar=3xx3%=9%` |
|
| 35. |
If the length of second pendulum becomes (1)/(3) what will be its periodic time? |
|
Answer» Solution :`implies (2)/(SQRT(3)) s` Time period of second PENDULUM `T_(1) = 2s` In EQUATION `T= 2pi sqrt((l)/(g)), 2pi, g` are constant `therefore T propto sqrt(l)` `therefore (T_2)/(T_1)= sqrt((l_2)/(l_1)) = sqrt((l)/(3XX l_(1)))= (1)/(sqrt(3))""[therefore l_(2)= (l_1)/(3)]` `therefore T_(2) = (T_1)/(sqrt(3))= (2)/(sqrt(3))s`. |
|
| 36. |
A ball is dropped from a bridge at a height of 176.4 m over a river. After 2s a second ball is thrown straight downwards. What should be the initial velocity of the second ball so that both hit the water simultaneously ? |
|
Answer» 2.45 m/s |
|
| 37. |
An artifical satellite of mass m is moving in a circular orbit at a height equal to the radius R of the earth. Suddenly due to internal explosion the satellite breaks into two parts of equal pieces. One part of the satellite stops just after the explosion. THe increase in the mechanical energy of the system due to explosion will be (given , acceleration due to gravity on the surface of earth is g ) |
|
Answer» MGR |
|
| 38. |
A body takes time to reach the bottom of an inclined plane of angle with the horizontal. If the plane is made rough, time taken now is 2t. The coefficient of friction of the rough surface is |
|
Answer» `3/4 TANTHETA ` |
|
| 39. |
Does the work done in raising a suitcase on to platform depend upon how far it is raised up? |
| Answer» SOLUTION :No, the work DONE does not DEPEND on the TIME rate with which the suitcase is RAISED. | |
| 40. |
A simple pendulu is of length 50 cm. Find its time period and frequency of oscillation. (g = 9.8 m//s^(2)). |
|
Answer» 2.419 SEC, 0.8045Hz |
|
| 41. |
Figure shows two coherent sources S_1 and S_2which emit sound of wavelength lamda in phase. The separation between the sources is 3lamda. A circular wire at large radius is placed in such a way that S_1 S_2 lies in its plane and the middle point of S_1 S_2, is at the centre of the wire. Find the angular positions 0 on the wire for which constructive interference takes place. |
Answer»  `(S_1P)^2=(S_1x)^2+(Px)^`………1 `(S_2P)^2=(S_2M)^2+(PM)^2`……..2 from 1 and 2 `(S_1P)^2-(S_2P)^2=(Px)^2-(PM)^2` `=(1.5lamda+Rcostheta)^2-(Rcostheta-1.5lamda)^2` `=6lamdacostheta` `rarr (S_1P-S_2P)^2=(6lamdaRcostheta)/(2R)` `=3costheta` For constructive interferene `(S_1P-S-2P)^2=x=3lamdacostheta=nlamda` `rarr costheta=n/3` `theta=cos^-1(n/3)` `where n=0,1,2,.............. `rarr theta=0^@, 48^@, 2^@, 7^@, 5^@, 90^@ and` SIMILAR POINTS in other QUADRANTS. |
|
| 42. |
When we try to close a water tap with our fingers, fastjets of water yush through the eyeing between our fingers why? |
| Answer» SOLUTION :. ACCORDING to equation of continuity, area `XX`velocity - a constant. Here are DECREASES, so velocity increases | |
| 43. |
When a specific gravity bottle filled with mercury at 0° is heated o 100°C, 5.190g of mercury overflows and 300g remains in the specific gravity bottle. Using glycerine in place of mercury, 1.439 of glycerine overflows and 27 .333 g remains. The coefficient of real expansion of mercury is 0.000183//^(@) C. then the coefficient of real expansion of glycerine is |
|
Answer» `5.4 XX 10^(-4)//^(@) C` |
|
| 44. |
A bullet of mass 10 gm hits a fixed target and penetrates through 8cm in it before coming to rest. If the average resistance offered by the target is 100 N, the velocity with which the bullet hits the target is, |
|
Answer» 20 m/s |
|
| 45. |
Two cannoballs A and B are fired at same angle and with same velocity from a cannon . If mass of A lt mass of B, which ball will reach the ground first ? What if the balls are fired with diferent velocities but same angle of projection? |
| Answer» SOLUTION :Time of flight of a projectileis directlyproportional to the initial velocity of projection . So, the BALL which is fired with lesservelocity will reach the GROUND earlier. | |
| 46. |
A bullet is fired from gun. The force on bullet is, F = 600 -2 xx 10^(5)newton. The force reduces to zero just when bullet leaves barrel. Find the impulse imparted to bullet |
|
Answer» 0.6NS |
|
| 47. |
A car is moving with a velocity of 20 m/s. The driver sees a stationary truck ahead at a distance of 100 m. After some reaction time Deltat the brakes are applied producing a retardation of 4 m//s^2? What is the maximum reaction time to avoid collision ? |
|
Answer» SOLUTION :After applyiing the breaks LET S be the distance covered by the car before coming to rest `v^(2)=U^(2)+2as` COVER distance s `0=20^(2)-2 xx 4s ""s=(400)/(8)=50m` The car covers 50m after applying breaks To AVOID the clash, the remaining distance 100-50=50m must be covered by the car with uniform velocity 20m/s during the reaction time `trianglet` `(50)/(trianglet)=20""trianglet=(50)/(20)=2.5s` The maximum reaction time `trianglet=2.5s` |
|
| 48. |
If the excess pressure inside a soap bubble of radius 5 mm is balanced by a 1.5 mm column of oil of specific gravity 0.76, find the surface tension of the soap solution. |
|
Answer» Solution :Given, `r=5 mm =5 xx 10^(-3) m` `h=1.5 mm =1.5 xx 10^(-3) m` `p_("oil")=0.76 xx 10^(3)" kg/m"^(3)` Excess pressure inside a soap bubble with two free surface is `p=(4 sigma)/R` which is EQUAL to the PRESURE EXERTED by 1.5 mm COLUMN of oi. `p=HPG` `rArr (4 sigma)/(R)=hpg` `or sigma=(hpgR)/(4)` `=(1.5 xx 10^(-3) xx 0.76 xx 10^(3) xx 9.8 xx 5 xx 10^(-3))/(4)` `=13.96 xx 10^(-3) N//m` |
|
| 49. |
The moment of inertia of a flywheel having kinetic energy 360 J and angular speed of 20 rad/s is |
|
Answer» 18 KG `m^(2)` Angular speed , `omega = 20` rad/s . As `K = (1)/(2) I omega^(2)` (where I = moment of inertia) `therefore I = (2K)/(omega^2) = (2 xx 360)/(20 xx 20) = 1.8 kg m^2` |
|
| 50. |
Figure shows the motion of a particle along a straight line. Find the average velocity of the particle during the intervals. (a) A to E, (b) B to E, (c ) C to E, (d) D to E, (e ) C to D |
Answer» Solution : (a) As the particle moves from A to E, A is the initial point and E is the final point. The slope of the line DRAWN from A to E. i.e, `(trianglex)/(trianglet)` gives the average velocity during that INTERVAL of time. The displacement `trianglex` is `x_(E)-x_(A)=10cm-0cm=+10cm` The time interval `trianglet_(EA)=t_(E)-t_(A)=10s`. During this interval average velocity `vecv=(trianglex)/(trianglet)=(+10cm)/(10s)=+1cms^(-1)` (b) During the interval B to E, the displacement `trianglex=x_(E)-x_(B)=10cm -4cm=6cm` and `trianglet=t_(E)-t_(B)=10s-3s=7s` Average VELOCTY `vecv=(trianglex)/(trianglet)=(6cm)/(7s)` `=+0.857 CMS^(-1) =0.86cms^(-1)` (c) During the interval C to E, the displacement `trianglex=x_(E)-x_(C)=10cm-12cm=2cm and ` `trianglet=t_(E)-t_(C)=10s-5s=5s` `vecv=(trianglex)/(trianglet)=(-2cm)/(5s)=-0.4cms^(-1)` (d) During the interval D to E, the displacement `trianglex=x_(E)-x_(D)=10cm-12cm=-2cm` and the time interval `trianglet=t_(E)-t_(D)=10s-8s=2s` ` vecv=(trianglex)/(trianglet)=(-2cm)/(2s)=-1cms^(-1)` (e) During the interval C to D, the displacement `trianglex-x_(D)-x_(C)=12cm ` 12cm=0 and the time interval. `triangle t=t_(D)-t_(C)=8s-5s=3s` The average velocity `vecv=(trianglex)/(trianglet)=(0m)/(3s) =0 ms^(-1)` (The particle has reached the same POSITION during these 3s. The average velocity is zero because the displacement is zero). |
|
