Objective question for Kinematics

1. A body is projected horizontally from a point above the ground.The motion of body is defined as
x=2t
y=2t2
where x and y are horizontal and vertical displacement respectivley at time t.Which one of the following is true
a.The trajectory of the body is a parabola
b The trajectory of the body is a straight line
c.the velocity vector at point t is 2i+4tj
d the acceleration vector at time t is 4j


2.when the projectile is at the highest point of its trajectory,the direction of its velocity and acceleration are
a. parallel to each other
b.antiparallel to each other
c. Inclined to each other at 45
d. Perpendicular to each other

3.The horizontal and vertical displacement of the projectile at time t are
x=36t
y=48t-4.9t2
where x and y are in meters and t in second.Intial velocity of the projectile in m/s
a. 15
c. 30
b. 45
d. 60

4.Displacement(y) of the particle is given by
y=2t+t2-2t2
The velocity of the particle when acceleration is zero is given by
a. 5/2
b. 9/4
c. 13/6
d. 17/8

5. A car ,starting from rest is accelerated at constant rate a until it attains speed V.It is then retarted at a constant rate b until it comes to rest.which of the following is true
a. the average speed for the whole motion is av/2b
b. the average speed for the whole motion is v/2
c. Total time taken for the journey is v(1/a+1/b)
d. none of the above

6.The intial velocty of the particle is u=4i+3j m/s.It is moving with uniform acceleration a=.4i+.3j m/s2.which of the following is true
a. the magnitude of the velocity after 10 sec is 10m/s
b. The velocity vector at time t is given by (4+.4t)i +(3+.3t)j
c. the displacement at time t is (4t+.22)i +(3t+.152)j
d . none of the above

7.It is possible to project an body with a given speed in two possible ways so that it has a same horizontal range.The product of time taken by it in tow possible ways is
a. R/g
b. 2R/g
c. 3R/g
d. 4R/g

8.A projectile has a range R and time of flight T.If the range is tripled by the increasing the speed of the projection ,without changing the angle of projection.The time of the flight will become
a T/√3
b T√3
c. T/3
d. 3T

9.A large number of stones are fired in all the direction with same speed V.The maximum area of the ground on which this stone will spread is
a. πV4/g2
b. π2V4/g2
c. πV4/g
d. πV/g2

10.A body is projected horizontally from the top of the building 39.2 m high.How long will it take to hit the ground
a. 2 sec
b. 4 sec
c. 1 sec
d 2√2 sec

Detailed Solutions

Ans 4:
given

y=2t+t2-2t3

Velocity is given
v=dy/dt=2+2t-6t2

a=dv/dt=2-12t

Now acceleratio is zero
2-12t=0
t=1/6

Putting this value velocity equation
v=2+2/6-6(1/6)2
=2+1/3-1/6
=13/6


Ans 5:

The distance s1 covered by the car during the time it is accelerated is given by
2as1=v2

Which gives
s1=v2/2a

Similary in the decelerated motion,distance covered is
2bs2=v2
Which gives
s2=v2/2b

SO Total distance travelled during the whole journey
s=s1+s2=v2/2(1/a+1/b)

Let t1 be the time taken during accelerated motion then
v=at1 or t1=v/a
and Let t2 be the time taken during deccelerated motion then
v=bt2 or t2=v/b

Total timetaken
t=t1+t2
=v(1/a+1/b)

So average speed =total distance/total time taken
=s/t=v/2





Ans 7:
If a body is projected with a given velocity u at angle θ and (90-θ) to the horizontal,it will have same range R given by
R=u2sin2θ/g

The corresponding times if flight are
t1=2usinθ/g
t2=2usin(90-θ)/g=2ucosθ/g
t1t2=2u2(2sinθcosθ)/g2
=2u2sin2θ/g2=2R/g

Ans 8

Range is given by
R=u2sin2θ/g-----(1)
and time taken by
T=2usinθ/g-----(2)

Now
3R=uc2sin2θ/g-----(3)
Tc=2ucsinθ/g-----(4)

Dividing 1 by 3
1/3=(u/uc)2 ----5
Dividing 2 by 4

T/Tc=u/uc ---6

From 5 and 6
T/Tc=1/√3
Tc=T/√3



Ans 10:

Intial vertical component of velocity is zero
so h=(1/2)gt2
39.2=(1/2)*9.8*t2
t=2√2 sec

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