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PHXI03:MOTION IN A STRAIGHT LINE

362332 A car is moving towards check post with velocity of $54 k m h^{- 1}$ . When the car is at a distance of 400 $m$ from the check post, the driver applies brakes which causes a deceleration of $0.3 {m s}^{- 2}$ . Find the distance of car from the check post for 2 minutes after applying the brakes.

1 25

2 33

3 20

4 39

Explanation:

Here, $u = 54 \times \frac{5}{18} = 15 {m s}^{- 1}$
$a = - 0.3 {m s}^{- 2}$
$\begin{array}{l} ∵ & v = u + a t \\ or & 0 = 15 - 0.3 t_{0} \end{array}$
$∴ t_{0} = 50 s$
After 50 s , car permanently comes to rest.
$∴ s = (\frac{u + v}{2}) t = (\frac{15 + 0}{2}) \times 50 = 375 m$
Thus, the distance of car from the check post is $(400 - 375) m = 25 m$

PHXI03:MOTION IN A STRAIGHT LINE

362333 A particle starts its motion from rest under the action of a constant force. If the distance covered in first $10 s i s s_{1}$ and that covered in the first $20 s i s s_{2}$ , then

1

s_{2} = 2 s_{1}

2

s_{2} = 3 s_{1}

3

s_{2} = s_{1}

4

s_{2} = 4 s_{1}

Explanation:

Since, the body starts from rest $u = 0$
$s = \frac{1}{2} a t^{2}$
Now, $s_{1} = \frac{1}{2} a (10)^{2}$ (1)
and $s_{2} = \frac{1}{2} a (20)^{2}$ (2)
Dividing Eq. (1) and Eq. (2), we get
$\frac{s_{1}}{s_{2}} = \frac{{(10)}^{2}}{{(20)}^{2}}$
$\Rightarrow s_{2} = 4 s_{1}$

PHXI03:MOTION IN A STRAIGHT LINE

362334 A particle start from rest with a velocity of $10 m / s$ and moves with a constant acceleration till the velocity increases to $100 m / s$ . At an instant the acceleration is simultaneously reversed, what will be the velocity of the particle when it comes back to the starting point?

1

10 m / s

2

20 m / s

3

30 m / s

4

40 m / s

Explanation:

According to relation $= v^{2} - u^{2} = 2 a s$
$(100)^{2} - (10)^{2} = 2 a s$
$10000 - 100 = 2 a s 9900 = 2 a s$
Now acceleration just reversed $= a = - a$
Particle comes back to original position $v =$ ?
$\begin{array}{ll} v^{2} - u^{2} = 2 a s & (v)^{2} - (100)^{2} = 2 (- a) s \\ v^{2} = 10000 - 2 a s & v^{2} = 10000 - 2 a s \end{array}$
Put 2 as $= 9900 v^{2} = 100 \Rightarrow v = \sqrt{100} = 10 m / s$

PHXI03:MOTION IN A STRAIGHT LINE

362335 A frictionless wire $A B$ is fixed on a sphere of radius $R$ . A very small spherial ball slips on this wire. The time taken by this ball to slip from $A$ to $B$ is
supporting img

1

2 \sqrt{\frac{R}{g}}

2

\frac{g R}{\sqrt{g \cos θ}}

3

\frac{2 \sqrt{g R}}{g \cos θ}

4

2 \sqrt{g R} . \frac{\cos θ}{g}

Explanation:

Acceleration of a body along $A B$ is $g \cos θ$
Distance travelled in time
$t \sec = A B = \frac{1}{2} (g \cos θ) t^{2}$
From
$Δ A B C, A B = 2 R \cos θ \Rightarrow 2 R \cos θ = \frac{1}{2} g \cos θ t^{2}$
$\Rightarrow t^{2} = \frac{4 R}{g} o r t = 2 \sqrt{\frac{R}{g}}$
Here ‘ $t$ ’ is independent of $^{'} θ^{'}$

PHXI03:MOTION IN A STRAIGHT LINE

362332 A car is moving towards check post with velocity of $54 k m h^{- 1}$ . When the car is at a distance of 400 $m$ from the check post, the driver applies brakes which causes a deceleration of $0.3 {m s}^{- 2}$ . Find the distance of car from the check post for 2 minutes after applying the brakes.

1 25

2 33

3 20

4 39

Explanation:

Here, $u = 54 \times \frac{5}{18} = 15 {m s}^{- 1}$
$a = - 0.3 {m s}^{- 2}$
$\begin{array}{l} ∵ & v = u + a t \\ or & 0 = 15 - 0.3 t_{0} \end{array}$
$∴ t_{0} = 50 s$
After 50 s , car permanently comes to rest.
$∴ s = (\frac{u + v}{2}) t = (\frac{15 + 0}{2}) \times 50 = 375 m$
Thus, the distance of car from the check post is $(400 - 375) m = 25 m$

PHXI03:MOTION IN A STRAIGHT LINE

362333 A particle starts its motion from rest under the action of a constant force. If the distance covered in first $10 s i s s_{1}$ and that covered in the first $20 s i s s_{2}$ , then

1

s_{2} = 2 s_{1}

2

s_{2} = 3 s_{1}

3

s_{2} = s_{1}

4

s_{2} = 4 s_{1}

Explanation:

Since, the body starts from rest $u = 0$
$s = \frac{1}{2} a t^{2}$
Now, $s_{1} = \frac{1}{2} a (10)^{2}$ (1)
and $s_{2} = \frac{1}{2} a (20)^{2}$ (2)
Dividing Eq. (1) and Eq. (2), we get
$\frac{s_{1}}{s_{2}} = \frac{{(10)}^{2}}{{(20)}^{2}}$
$\Rightarrow s_{2} = 4 s_{1}$

PHXI03:MOTION IN A STRAIGHT LINE

362334 A particle start from rest with a velocity of $10 m / s$ and moves with a constant acceleration till the velocity increases to $100 m / s$ . At an instant the acceleration is simultaneously reversed, what will be the velocity of the particle when it comes back to the starting point?

1

10 m / s

2

20 m / s

3

30 m / s

4

40 m / s

Explanation:

According to relation $= v^{2} - u^{2} = 2 a s$
$(100)^{2} - (10)^{2} = 2 a s$
$10000 - 100 = 2 a s 9900 = 2 a s$
Now acceleration just reversed $= a = - a$
Particle comes back to original position $v =$ ?
$\begin{array}{ll} v^{2} - u^{2} = 2 a s & (v)^{2} - (100)^{2} = 2 (- a) s \\ v^{2} = 10000 - 2 a s & v^{2} = 10000 - 2 a s \end{array}$
Put 2 as $= 9900 v^{2} = 100 \Rightarrow v = \sqrt{100} = 10 m / s$

PHXI03:MOTION IN A STRAIGHT LINE

362335 A frictionless wire $A B$ is fixed on a sphere of radius $R$ . A very small spherial ball slips on this wire. The time taken by this ball to slip from $A$ to $B$ is
supporting img

1

2 \sqrt{\frac{R}{g}}

2

\frac{g R}{\sqrt{g \cos θ}}

3

\frac{2 \sqrt{g R}}{g \cos θ}

4

2 \sqrt{g R} . \frac{\cos θ}{g}

Explanation:

Acceleration of a body along $A B$ is $g \cos θ$
Distance travelled in time
$t \sec = A B = \frac{1}{2} (g \cos θ) t^{2}$
From
$Δ A B C, A B = 2 R \cos θ \Rightarrow 2 R \cos θ = \frac{1}{2} g \cos θ t^{2}$
$\Rightarrow t^{2} = \frac{4 R}{g} o r t = 2 \sqrt{\frac{R}{g}}$
Here ‘ $t$ ’ is independent of $^{'} θ^{'}$

PHXI03:MOTION IN A STRAIGHT LINE

362332 A car is moving towards check post with velocity of $54 k m h^{- 1}$ . When the car is at a distance of 400 $m$ from the check post, the driver applies brakes which causes a deceleration of $0.3 {m s}^{- 2}$ . Find the distance of car from the check post for 2 minutes after applying the brakes.

1 25

2 33

3 20

4 39

Explanation:

Here, $u = 54 \times \frac{5}{18} = 15 {m s}^{- 1}$
$a = - 0.3 {m s}^{- 2}$
$\begin{array}{l} ∵ & v = u + a t \\ or & 0 = 15 - 0.3 t_{0} \end{array}$
$∴ t_{0} = 50 s$
After 50 s , car permanently comes to rest.
$∴ s = (\frac{u + v}{2}) t = (\frac{15 + 0}{2}) \times 50 = 375 m$
Thus, the distance of car from the check post is $(400 - 375) m = 25 m$

PHXI03:MOTION IN A STRAIGHT LINE

362333 A particle starts its motion from rest under the action of a constant force. If the distance covered in first $10 s i s s_{1}$ and that covered in the first $20 s i s s_{2}$ , then

1

s_{2} = 2 s_{1}

2

s_{2} = 3 s_{1}

3

s_{2} = s_{1}

4

s_{2} = 4 s_{1}

Explanation:

Since, the body starts from rest $u = 0$
$s = \frac{1}{2} a t^{2}$
Now, $s_{1} = \frac{1}{2} a (10)^{2}$ (1)
and $s_{2} = \frac{1}{2} a (20)^{2}$ (2)
Dividing Eq. (1) and Eq. (2), we get
$\frac{s_{1}}{s_{2}} = \frac{{(10)}^{2}}{{(20)}^{2}}$
$\Rightarrow s_{2} = 4 s_{1}$

PHXI03:MOTION IN A STRAIGHT LINE

362334 A particle start from rest with a velocity of $10 m / s$ and moves with a constant acceleration till the velocity increases to $100 m / s$ . At an instant the acceleration is simultaneously reversed, what will be the velocity of the particle when it comes back to the starting point?

1

10 m / s

2

20 m / s

3

30 m / s

4

40 m / s

Explanation:

According to relation $= v^{2} - u^{2} = 2 a s$
$(100)^{2} - (10)^{2} = 2 a s$
$10000 - 100 = 2 a s 9900 = 2 a s$
Now acceleration just reversed $= a = - a$
Particle comes back to original position $v =$ ?
$\begin{array}{ll} v^{2} - u^{2} = 2 a s & (v)^{2} - (100)^{2} = 2 (- a) s \\ v^{2} = 10000 - 2 a s & v^{2} = 10000 - 2 a s \end{array}$
Put 2 as $= 9900 v^{2} = 100 \Rightarrow v = \sqrt{100} = 10 m / s$

PHXI03:MOTION IN A STRAIGHT LINE

362335 A frictionless wire $A B$ is fixed on a sphere of radius $R$ . A very small spherial ball slips on this wire. The time taken by this ball to slip from $A$ to $B$ is
supporting img

1

2 \sqrt{\frac{R}{g}}

2

\frac{g R}{\sqrt{g \cos θ}}

3

\frac{2 \sqrt{g R}}{g \cos θ}

4

2 \sqrt{g R} . \frac{\cos θ}{g}

Explanation:

Acceleration of a body along $A B$ is $g \cos θ$
Distance travelled in time
$t \sec = A B = \frac{1}{2} (g \cos θ) t^{2}$
From
$Δ A B C, A B = 2 R \cos θ \Rightarrow 2 R \cos θ = \frac{1}{2} g \cos θ t^{2}$
$\Rightarrow t^{2} = \frac{4 R}{g} o r t = 2 \sqrt{\frac{R}{g}}$
Here ‘ $t$ ’ is independent of $^{'} θ^{'}$

PHXI03:MOTION IN A STRAIGHT LINE

362332 A car is moving towards check post with velocity of $54 k m h^{- 1}$ . When the car is at a distance of 400 $m$ from the check post, the driver applies brakes which causes a deceleration of $0.3 {m s}^{- 2}$ . Find the distance of car from the check post for 2 minutes after applying the brakes.

1 25

2 33

3 20

4 39

Explanation:

Here, $u = 54 \times \frac{5}{18} = 15 {m s}^{- 1}$
$a = - 0.3 {m s}^{- 2}$
$\begin{array}{l} ∵ & v = u + a t \\ or & 0 = 15 - 0.3 t_{0} \end{array}$
$∴ t_{0} = 50 s$
After 50 s , car permanently comes to rest.
$∴ s = (\frac{u + v}{2}) t = (\frac{15 + 0}{2}) \times 50 = 375 m$
Thus, the distance of car from the check post is $(400 - 375) m = 25 m$

PHXI03:MOTION IN A STRAIGHT LINE

362333 A particle starts its motion from rest under the action of a constant force. If the distance covered in first $10 s i s s_{1}$ and that covered in the first $20 s i s s_{2}$ , then

1

s_{2} = 2 s_{1}

2

s_{2} = 3 s_{1}

3

s_{2} = s_{1}

4

s_{2} = 4 s_{1}

Explanation:

Since, the body starts from rest $u = 0$
$s = \frac{1}{2} a t^{2}$
Now, $s_{1} = \frac{1}{2} a (10)^{2}$ (1)
and $s_{2} = \frac{1}{2} a (20)^{2}$ (2)
Dividing Eq. (1) and Eq. (2), we get
$\frac{s_{1}}{s_{2}} = \frac{{(10)}^{2}}{{(20)}^{2}}$
$\Rightarrow s_{2} = 4 s_{1}$

PHXI03:MOTION IN A STRAIGHT LINE

362334 A particle start from rest with a velocity of $10 m / s$ and moves with a constant acceleration till the velocity increases to $100 m / s$ . At an instant the acceleration is simultaneously reversed, what will be the velocity of the particle when it comes back to the starting point?

1

10 m / s

2

20 m / s

3

30 m / s

4

40 m / s

Explanation:

According to relation $= v^{2} - u^{2} = 2 a s$
$(100)^{2} - (10)^{2} = 2 a s$
$10000 - 100 = 2 a s 9900 = 2 a s$
Now acceleration just reversed $= a = - a$
Particle comes back to original position $v =$ ?
$\begin{array}{ll} v^{2} - u^{2} = 2 a s & (v)^{2} - (100)^{2} = 2 (- a) s \\ v^{2} = 10000 - 2 a s & v^{2} = 10000 - 2 a s \end{array}$
Put 2 as $= 9900 v^{2} = 100 \Rightarrow v = \sqrt{100} = 10 m / s$

PHXI03:MOTION IN A STRAIGHT LINE

362335 A frictionless wire $A B$ is fixed on a sphere of radius $R$ . A very small spherial ball slips on this wire. The time taken by this ball to slip from $A$ to $B$ is
supporting img

1

2 \sqrt{\frac{R}{g}}

2

\frac{g R}{\sqrt{g \cos θ}}

3

\frac{2 \sqrt{g R}}{g \cos θ}

4

2 \sqrt{g R} . \frac{\cos θ}{g}

Explanation:

Acceleration of a body along $A B$ is $g \cos θ$
Distance travelled in time
$t \sec = A B = \frac{1}{2} (g \cos θ) t^{2}$
From
$Δ A B C, A B = 2 R \cos θ \Rightarrow 2 R \cos θ = \frac{1}{2} g \cos θ t^{2}$
$\Rightarrow t^{2} = \frac{4 R}{g} o r t = 2 \sqrt{\frac{R}{g}}$
Here ‘ $t$ ’ is independent of $^{'} θ^{'}$

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