# Motion In A Straight Line - Class 11 - Physics

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## Extra Questions - Motion In A Straight Line | NCERT | Physics | Class 11

A particle accelerates from rest at a constant rate for some time and attains a peak velocity of $8 \mathrm{~m} \mathrm{~s}^{-1}$. Afterwards, it decelerates at the same rate and comes to rest. If the total time taken is 4 seconds, the total distance traveled is $\qquad$

A) $32 \mathrm{~m}$

B) $16 \mathrm{~m}$

C) $4 \mathrm{~m}$

D) $25 \mathrm{~m}$

The correct answer is **Option B: $16 , \text{m}$**.

To solve this problem, visualize it using a **velocity-time graph**, where the particle starts from rest, accelerates to a peak velocity, then decelerates back to rest.

Given:

The peak velocity $ v = 8 , \text{m/s} $

The total time $ t = 4 , \text{s} $

Since the acceleration and deceleration are uniform and equal in magnitude, the graph is symmetric, and the peak velocity is reached at the midpoint of the total time. Therefore, the time to reach the peak velocity is $ t/2 = 4/2 = 2 , \text{s} $.

The graph forms a triangle under the velocity curve. The area under this triangle (which represents the distance traveled) can be calculated using the formula for the area of a triangle:

$$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 8 $$

Calculating this gives:

$$ \text{Area} = \frac{1}{2} \times 4 \times 8 = 16 , \text{m} $$

Thus, the total distance traveled by the particle is **$16 , \text{m}$**.

If the velocity of a particle is $v(t) = 2t - 4 , \mathrm{m/s}$, the average speed and magnitude of average velocity of the particle in $10$ seconds respectively are

A) $4.8 , \mathrm{m/s}, , 4 , \mathrm{m/s}$

B) $6 , \mathrm{m/s}, , 6 , \mathrm{m/s}$

C) $6.8 , \mathrm{m/s}, , 6 , \mathrm{m/s}$

D) $5.2 , \mathrm{m/s}, , 4.4 , \mathrm{m/s}$

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Two trains moving in the same direction with the same speed will be:

A. At rest with respect to each other.

B. In motion with respect to each other.

C. At rest with respect to an observer standing outside the train.

D. In motion with respect to an observer standing outside the train.

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A ball is dropped from a pole of height 100 m and simultaneously a stone is thrown upwards; they meet at a point 55 m above the ground. The velocity with which the stone must be thrown should equal:

A) 100/3 m/s

B) 200/3 m/s

C) 100 m/s

D) 400/3 m/s

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A block is thrown with a velocity of $2 \mathrm{~ms}^{-1}$ (relative to ground) on a belt, which is moving with velocity $4 \mathrm{~ms}^{-1}$ in the opposite direction of the initial velocity of the block. If the block stops slipping on the belt after $4 \mathrm{~s}$ after it is thrown, then choose the correct statement(s). Assume constant deceleration of the block.

Which of the following statements are correct?

A. Displacement with respect to ground is zero after $2.66 \mathrm{~s}$ and the magnitude of displacement with respect to the ground is $12 \mathrm{~m}$ after $4 \mathrm{~s}$.

B. The magnitude of displacement with respect to the ground in $4 \mathrm{~s}$ is $4 \mathrm{~m}$.

C. The magnitude of displacement with respect to the belt in $4 \mathrm{~s}$ is $12 \mathrm{~m}$.

D. Displacement with respect to the ground is zero in $8/3 \mathrm{~s}$.

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Which of the following statements is/are not true for a round trip along the same route?

A. Average speed is zero in a round trip.

B. Average velocity is zero in a round trip.

C. Average velocity and average speed, both are non-zero.

D. Average velocity is non-zero because displacement is non-zero.

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Answer:

A motorcycle moving with a speed of $5 , \mathrm{m/s}$ is subjected to an acceleration of $0.2 , \mathrm{m/s}^2$. We can use the following kinematic equations to calculate the speed of the motorcycle after 10 seconds and the distance traveled in this time:

- Speed after time $t$: $$ v = u + at $$ where: $v$ = final speed, $u$ = initial speed, $a$ = acceleration, and $t$ = time.

Substitute the values: $v = 5 , \mathrm{m/s} + (0.2 , \mathrm{m/s}^2) \times 10 , \mathrm{s} = 7 , \mathrm{m/s}$

Thus, the speed of the motorcycle after 10 seconds is $7 , \mathrm{m/s}$.

- Distance traveled after time $t$: $$ s = ut + \frac{1}{2}at^2 $$ where: $s$ = distance traveled.

Substitute the values: $s = (5 , \mathrm{m/s} \times 10 , \mathrm{s}) + \frac{1}{2} \times (0.2 , \mathrm{m/s}^2) \times (10 , \mathrm{s})^2 = 60 , \mathrm{m}$

Therefore, the distance traveled by the motorcycle in 10 seconds is 60 meters.

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A man leaves his house for a cycle ride. He comes back to his house after half an hour after covering a distance of one kilometer. What is his average velocity for the ride?

A) Zero

B) $2 \mathrm{~km} \mathrm{~h}^{-1}$

C) $10 \mathrm{~km} \mathrm{~h}^{-1}$

D) $0.5 \mathrm{~km} \mathrm{~h}^{-1}$

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The rear side of a truck is open and a box of $40 \mathrm{~kg}$ mass is placed $5 \mathrm{~m}$ away from the open end as shown in Fig. 5.22. The coefficient of friction between the box and the surface below it is $0.15$. On a straight road, the truck starts from rest and accelerates with $2 \mathrm{~ms}^{-2}$. At what distance from the starting point does the box fall off the truck?