- 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
- |
- Velocity and acceleration
- |
- Uniformly accelerated motion
- |
- Relative Velocity
- |
- Kinematics Question 1
- |
- Kinematics Question 2

In this page we have *Acceleration problems with solutions* . Hope you like them and do not forget to like , social share
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**13.** 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(13)**:

. 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

**14.**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(14)**:

. 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

**15.**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(15)**:

. Here the nut intially 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

**16.** Let v and a denote the velocity and acceleration respectivly 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(16)**:

. 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 ,accleration 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

**More examples**

**Question 1** A Stone is thrown Up, what is the velocity, acceleration at the top and what direction?

**Solution**

At the peak,velocity of the stone is zero but accleration is not zero, acceleration is equal to acceleration due to gravity and it is acting downwards

**Question 2**What does the slope of the velocity-time graph shows?

**Solution**

Acceleration

**Question 3**

What does the area under the velocity time graph represent

**Solution**

Displacement

**Question 4**

Give example of the motion where x>0, v < 0 and a > 0?

**Solution**

x= p +qe^{-t} Where p and q > 0

So x > 0 at any instant

Velocity(dx/dt) = d(p +qe^{-t})/dt= -qe^{-t} ( v < 0)

Acceleration(dv/dt) = d(-qe^{-t})/dt= qe^{-t} ( a > 0)

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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. 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. 13.6 m

b. 10 m

c. 12.6 m

d. none of these

. Here the nut intially 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 ,accleration 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

At the peak,velocity of the stone is zero but accleration is not zero, acceleration is equal to acceleration due to gravity and it is acting downwards

Acceleration

What does the area under the velocity time graph represent

Displacement

Give example of the motion where x>0, v < 0 and a > 0?

x= p +qe

So x > 0 at any instant

Velocity(dx/dt) = d(p +qe

Acceleration(dv/dt) = d(-qe

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