- Introduction
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- Position and Displacement
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- Average velocity and speed
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- Instantaneous velocity and speed
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- Acceleration
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- Motion with constant acceleration
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- Free fall acceleration
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- Relative velocity
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- Solved Examples Part 1
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- Solved Examples Part 2
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- Solved Examples Part 3
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- Solved Examples Part 4
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- Solved Examples Part 5

- Position Distance and Displacement
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- Average velocity and speed
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- Velocity and acceleration
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- Uniformly accelerated motion
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- Relative Velocity
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- Kinematics Question 1
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- Kinematics Question 2

In this page we have *Problems for Motion with Constant Acceleration in One Dimension* . Hope you like them and do not forget to like , social share
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**Question 9.**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. (ab/a+b)t

b.(a+b/ab)t

c. (a^{2}+b^{2}/ab)t

d.(a^{2}-b^{2}/ab)t

**Solution(9)**:

. Let t_{1} and t_{2} be the the time for acceleration and deccleration.

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=v/a + v/b

or v=abt/(a+b)

Hence (a) is correct

**Question 10.**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(10)**:

. given

y=2t+t^{2}-2t^{3}

Velocity is given

v=dy/dt=2+2t-6t^{2}

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

Hence (c) is correct

**Question 11.**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(11)**:

. Take the case of uniform circular motion,Instantanous 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

**Matrix match type**

**Question 12.**In a free fall motion from rest,Match column I to column II

**column I**

A) Graph between displacement and time

B) Graph between velocity and time

C) Graph between velocity and displacement

D) Graph between KE and displacement

**column II**

P) Parabola

Q) Straight line

C) Circle

D) No appropiate match given

**Solution(12)**:

. 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

**More Practice Questions**

**Question 1**

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 2**

A car moving with a speed of 40km/h can be stopped by applying the brakes after atleast 2 m.If the same car is moving with the speed 80K/h,what is the minimum stopping distance?

**Solution**

8 m

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a. (ab/a+b)t

b.(a+b/ab)t

c. (a

d.(a

. Let t

Let v be the maximum velocity attained

Then

v=at

v=bt

Now t=t

or t=v/a + v/b

or v=abt/(a+b)

Hence (a) is correct

y=2t+t

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

Velocity is given

v=dy/dt=2+2t-6t

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+1/3-1/6

=13/6

Hence (c) is correct

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,Instantanous 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 appropiate 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 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 40km/h can be stopped by applying the brakes after atleast 2 m.If the same car is moving with the speed 80K/h,what is the minimum stopping distance?

8 m

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