In this page we have 1D Kinematics Problems And Solutions . Hope you like them and do not forget to like , social share
and comment at the end of the page.
Multiple Choice questions
Question 1.
A truck accelerates from rest at the constant rate 'a' for some time after which it decelerates at a constant rate of 'b' to come to the rest.If the total time elapsed is t ,then find out the maximum velocity attains by the truck
a.$(\frac {ab}{a+b})t$
b.$(\frac {a+b}{ab})t$
c.$( \frac {a^2 +b^2}{ab})t$
d.$( \frac {a^2-b^2}{ab})t$ Solution:
Let t_{1} and t_{2} be the the time for acceleration and deceleration.
Let v be the maximum velocity attained
Then
v=at_{1} or t_{1}=v/a
v=bt_{2} or t_{2}=v/b
Now t=t_{1} + t_{2}
or $t= \frac {v}{a} + \frac {v}{b}$
or $v=\frac {abt}{a+b}$
Hence (a) is correct
Question 2.
Displacement(y) of the particle is given by
$y=2t+t^2-2t^3$
The velocity of the particle when acceleration is zero is given by
a. 5/2
b. 9/4
c. 13/6
d. 17/8 Solution:
given
$y=2t+t^2-2t^3$
Velocity is given
$v= \frac {dy}{dt}=2+2t-6t^2$
$a=\frac {dv}{dt}=2-12t$
Now acceleration 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
Hence (c) is correct
Question 3.
Mark out the correct statement
a. Instantaneous Velocity vector is always in the direction of the motion
b. Instantaneous acceleration vector is always in the direction of the motion
c. Acceleration of the moving particle can change its direction without any change in direction of velocity
d. None of the above Solution:
Take the case of uniform circular motion,Instantaneous Velocity vector and acceleration vector at any point is tangent and radial to the circle.So it is not along the direction of the circle
Take the case the moving car in one direction.If the car accelerated,acceleration is along the direction of velocity.if car driver put a brake then it deacclerates without any change in the direction of the velocity
Hence (c) is correct
Question 4.
In a free fall motion from rest,Match column I to column II
Solution:
Equation of motion for a free fall from rest
x=(1/2)gt^{2}.It is a parabola
v=gt it is a straight line
v^{2}=2gx it is a parabola
KE=(1/2)mv^{2}=mgx..... it is a straight line
So the correct answers are
A -> P
B -> Q
C -> P
D -> Q
Question 5
A body fall from height H.if t_{1} is time taken for covering first half height and t_{2} be time taken for second half.Which of these relation is true for t_{1} and t_{2}
a. t_{1} > t_{2}
b. t_{1} < t_{2}
c. t_{1}=t_{2}
d. Depends on the mass of the body Solution(5):
Let H be the height
then
First Half
H/2=(1/2)gt_{1}^{2} ----(1)
or
(1/2)gt_{1}^{2}=H/2
Also v=gt_{1}
Second Half
H/2=vt_{2}+(1/2)gt_{2}^{2}
or
(H/2)=gt_{1}t_{2}+(1/2)gt_{2}^{2}
or
(1/2)gt_{2}^{2} =(H/2)-gt_{1}t_{2} ---(2)
From 1 and 2
(1/2)gt_{2}^{2}=(1/2)gt_{1}^{2} -gt_{1}t_{2}
t_{2}^{2}+2t_{1}t_{2} -t_{1}^{2}=0
or t_{2}=[-2t_{1}+√(4t_{1}t_{2} +4t_{1}t_{2} )]/2
or t_{2}=[-2t_{1}+2t_{1}√8]/2
t_{2}=.44t_{1}
so t_{1} > t_{2}
Hence a is correct
Question 6
A train running at 30m/s is slowed uniformly to a stop in 44 sec.Find the stopping distance?
a. 612 m
b. 662 m
c. 630 m
d. 605 m
Solution(6):
Here
v_{0}=30m/s v_{f}=0 at t=44 sec
v_{f}=v_{0}+at or 0=30+a(44) or a=-.68m/s^{2}
Now x=v_{0}t+(1/2)at^{2}
or x=662 m
Hence b is correct
Question 7
A nut comes loose from a bolt on the bottom of an elevator as the elavator is moving up the shaft at 3m/s.The nut strikes the bottom of the shaft in 2 sec.How far from the bottom of the shaft was the elevator when nut falls off?
a. 13.6 m
b. 10 m
c. 12.6 m
d. none of these
Solution:
Here the nut initially has the velocity of the elevator ,so choosing upward as positive ,v_{0}=3 m/s
also a=-g=-9.8 m/s^{2}
The time taken to hit the bottom is 2 s so
y=v_{0}t+(1/2)at^{2}
=-13.6 m
Hence the bottom of the shaft was 13.6 m below elevator when nut fall off
Hence (a) is correct
Question 8
Let v and a denote the velocity and acceleration respectively of the particle in one dimension motion.
a.the speed of the particle decreases when v.a <0
b. the speed of the particle decreases when v.a >0
c.the speed of the particle increases when v.a=0
d. the speed of the particle decreases when |v|<|a| Solution:
when v.a <0
That mean they are in opposite direction
So when the particle is moving towards origin,acceleration is acting outwards so it is decreasing the speed
When the particle is moving outwards,acceleration is acting inwards hence it is decreasing the speed
So this option is correct
When v.a > 0
That means v and a are in same direction
When the particle is moving towards origin, acceleration is also acting inwards so increasing the speed
When the particle is moving outwards,acceleration is also acting outwards hence increasing the speed
So this option is not correct
when v.a=0
Then possible cases are
Acceleration is zero
Velocity is zero
Both are zero
So speed does not increases in all cases.So this option is not correct
When
|v|<|a|
Possible cases
Velocity is positive ,acceleration is positive ----Speed increase
Velocity is negative ,acceleration is negative----Speed increase
Velocity is positive,acceleration is negative
Velocity is negative ,acceleration is positive---Speed decreases
So this is not correct option
So correct answer is a
Watch this tutorial for learn about solving Kinematics equation Problems
Question 9
The displacement of a particle moving in straight line depends on time t as
x=at^{3}+bt^{2}+ct+d
which of the following is true
a. Initial acceleration depends on b only
b Initial velocity depends on c only
c. Initial displacement is d
d. Ratio of initial velocity /initial acceleration depends on a and c Solution:
The displacement of a particle moving in straight line depends on time t as
x=at^{3}+bt^{2}+ct+d
Question 10
A particle located at x=0 at time t=0 starts moving along the positive x-direction with a velocity v that varies as
v=√x
The displacement of the particles varies with time as
which of the following is true
a. t^{2}
b t^{3}
c. t^{4}
d. t^{1/2} Solution:
Given
v=√x
or
dx/dt=√x
dx/√x=dt
Integrating both sides between the limit (0,x) and (0,t)
x=t^{2}/4
Hence (a) is correct
Linked Answer type questions
Question 11
A Train is moving along a straight section of the track with a velocity of 180km/h. The braking deceleration is 2m/s^{2}s
At what distance from a train station should the train driver aply the brake so that train stops at the station
a. 800m
b 625m
c. 700m
d. none of these Solution:
we are given
v_{0}=180km/h=50m/s
a=-2m/s^{2}
Now
v^{2}=v_{0}^{2}+2as
0=(50)^{2}-2*2*x
or x=625m
hence (b) is correct
Question 12
from the previous question,how long will it take to bring the train to the halt
a. 25s
b 20s
c. 15s
d. none of these Solution(12):
Now
v=v_{0}+at
0=50-2t
or t=25 sec
Hence (a) is correct
Watch this tutorial for learn about solving Kinematics equation Problems
Long answer type questions
Question 13
A particle moves along the x axis. Its position id given by the equation x = 2 + 3t - 4t^{2}, with x in metres and t in seconds. Determine
(a) its position when it changes direction
(b) its velocity when it returns to the position it had at time t = 0. Solution
(a) x = 2 + 3t - 4t^{2}
velocity (v) = dx/dt = 3-8t
Change in Direction will happen when velocity becomes zero and start becoming negative
So 3-8t=0
t= 3/8 sec
Position will be given as
x= 2+ 3(3/8) -4(3/8)^{2}
x= 41/16 m
(b) Position at t=0
x= 2 m
Now we need to find what all time x=2
So
2== 2 + 3t ? 4t^{2}
t(3-4t)=0
t=0 or t =3/4
Now velocity at t=3/4
velocity (v) = dx/dt = 3-8t
= 3-8(3/4)=1 m/s
Question 14
A car moving with a speed of 40 km/h can be stopped by applying the brakes after at-least 2 m.If the same car is moving with the speed 80K/h,what is the minimum stopping distance? Solution
$v^2 =u^2+2as$
Applying for first, we get a=-400 m/s^{2}
Applying for second, we get
s=8 m
Question 15
The motion of a particle is described by the equation u=at.The distance travelled by the particle in first 4 sec is? Solution
8a
Question 16
Three different objects of masses 5 kg, 6 kg and 7 kg are allowed to fall from the same point O along three different frictionless path.What is the ratio of the speeds of the objects on reaching the grounds Solution
1:1:1
As acceleration is same
Question 17
A truck moves from point A to B with a uniform speed a and returns from B to A with the uniform speed of b. What is the average speed of the round trip Solution
2ab/(a+b)
link to this page by copying the following text Also Read