- distance and displacement
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- Velocity and speed
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- Instantaneous velocity and speed
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- Acceleration
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- Kinematic equations for uniformly accelerated Motion
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- Free fall acceleration
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- Relative velocity
- |
- Kinematics Sample Problems and Solutions
- |
- One dimensional motion problems with solution
- |
- Motion graphs worksheet with Answer

- Important Questions on Kinematics
- |
- Projectile motion problems
- |
- Kinematics and Projectile motion worksheet
- |
- Motion in one dimension (obj)
- |
- Motion in one dimension (sub)
- |
- Motion in one dimension Practice Paper
- |
- Acceleration worksheet with answers

In this page we have *1D Kinematics Sample Problems And Solutions* . Hope you like them and do not forget to like , social share
and comment at the end of the page.

a.$(\frac {ab}{a+b})t$

b.$(\frac {a+b}{ab})t$

c.$( \frac {a^2 +b^2}{ab})t$

d.$( \frac {a^2-b^2}{ab})t$

. Let t

Let v be the maximum velocity attained

Then

v=at

v=bt

Now t=t

or $t= \frac {v}{a} + \frac {v}{b}$

or $v=\frac {abt}{a+b}$

Hence (a) is correct

$y=2t+t^2-2t^3$

The velocity of the particle when acceleration is zero is given by

a. 5/2

b. 9/4

c. 13/6

d. 17/8

. 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+1/3-1/6

=13/6

Hence (c) is correct

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

. 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

A) Graph between displacement and time

B) Graph between velocity and time

C) Graph between velocity and displacement

D) Graph between KE and displacement

P) Parabola

Q) Straight line

C) Circle

D) No appropriate match given

. Equation of motion for a free fall from rest

x=(1/2)gt

v=gt it is a straight line

v

KE=(1/2)mv

So the correct answers are A -> P

B -> Q

C -> P

D -> Q

A body fall from height H.if t

t

a. t

b. t

c. t

d. Depends on the mass of the body

. Let H be the height

then

First Half

H/2=(1/2)gt

or

(1/2)gt

Also v=gt

Second Half

H/2=vt

or

(H/2)=gt

or

(1/2)gt

From 1 and 2

(1/2)gt

t

or t

or t

t

so t

Hence a is correct

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

. Here

v

v

Now x=v

or x=662 m

Hence b is correct

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

. Here the nut initially has the velocity of the elevator ,so choosing upward as positive ,v

also a=-g=-9.8 m/s

The time taken to hit the bottom is 2 s so

y=v

=-13.6 m

Hence the bottom of the shaft was 13.6 m below elevator when nut fall off

Hence (a) is correct

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|

. 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

The displacement of a particle moving in straight line depends on time t as

x=at

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

. The displacement of a particle moving in straight line depends on time t as

x=at

Velocity (dx/dt)=3at

Acceleration (d

So Initial displacement(t=0) =d

Initial velocity(t=0) =c

Initial acceleration (t=0) =2b

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

b t

c. t

d. t

. Given v=√x

or

dx/dt=√x

dx/√x=dt

Integrating both sides between the limit (0,x) and (0,t)

x=t

Hence (a) is correct

A Train is moving along a straight section of the track with a velocity of 180km/h. The braking deceleration is 2m/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

. we are given

v

a=-2m/s

Now v

0=(50)

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

. Now v=v

0=50-2t

or t=25 sec

Hence (a) is correct

A particle moves along the x axis. Its position id given by the equation x = 2 + 3t - 4t

(a) its position when it changes direction

(b) its velocity when it returns to the position it had at time t = 0.

a) x = 2 + 3t - 4t

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)

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

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

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?

8 m

The motion of a particle is described by the equation u=at.The distance travelled by the particle in first 4 sec is?

8a

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

1:1:1

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

2ab/(a+b)

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