Average Speed and Velocity - NEET Physics Questions

Question 1:

A particle is moving with a constant speed v on a circle of radius R, then find average acceleration of particle during half cycle is :

1. \[\frac{2\sqrt{2}v^{2}}{\pi R}\]

2. \[\frac{2v^{2}}{\pi R}\]

3. \[\frac{2v^{2}}{R}\]

4. Zero

View Answer

Change in velocity = 2V

Time taken = πR/V so,

Average acceleration = 2V/(πR/V)

Question 2:

A car is moving along a straight road with a uniform acceleration. It passes through two points P and Q separated by a distance with velocity 30 km/hr and 40 km/hr respectively. The velocity of the car midway between P and Q is :

1. 33.3 km/hr

2. 20√3 km/hr

3. 25√2 km/hr

4. 35 km/hr

View Answer

Speed at mid point is given by

\[ V_{mid}=\sqrt{\frac{V_{1}^{2}+V_{2}^{2}}{2}} \]

Question 3:

Between two stations, a train accelerates from rest uniformly at first, then moves with constant velocity, and finally retards uniformly to come to rest. If the ratio of the time taken is 1 : 8 : 1 and the maximum speed attained be 60 km h–¹, then what is the average speed over the whole journey?

1. \[48 km h^{-1}\]

2. \[52 km h^{-1}\]

3. \[54 km h^{-1}\]

4. \[56 km h^{-1}\]

View Answer

Time Ratios: The time ratios are given as $1:8:1$

$1 : 8 : 1$ . So, if the total time is $t$

$t$ , the train spends $\frac{t}{10}$

$10 t$ accelerating, $\frac{8t}{10}$

$10 8 t$ at constant velocity, and $\frac{t}{10}$

$10 t$ decelerating.
Velocity-Time Graph:
- The graph forms a trapezium.
- The train accelerates linearly from 0 to 60 km/h, holds constant at 60 km/h, and then decelerates back to 0.
Key points:
- The area under this graph represents the total distance traveled.
- The height (maximum velocity) = 60 km/h.
- The time intervals are in the ratio $1:8:1$
  
  $1 : 8 : 1$ .
Average Speed:
The area of the trapezium is given by:

$\text{Area} = \frac{1}{2} \times (t_{\text{total}}) \times (\text{initial velocity} + \text{final velocity})$ $Area = 21 \times (t_{total}) \times (initial velocity + final velocity)$ For the constant velocity portion:

$\text{Average speed} = \frac{60 \times (1 + 8 + 1)}{10} = 54 \, \text{km/h}$ $Average speed = 1060 \times ( 1 + 8 + 1 ) = 54 km/h$

Thus, the average speed is 54 km/h.

Question 4:

If velocity of a particle is given by V = (t + 3) m/s, then average velocity in interval

0 ≤ t ≤ 1s is :

1. 7/2 m/s

2. 9/2 m/s

3. 5 m/s

4. 4 m/s

View Answer

\[ V_{av}=\frac{\int_{0}^{1}v.dt}{\int_{0}^{1}dt}= \frac{\int_{0}^{1}(t + 3).dt}{\int_{0}^{1}dt}=\frac{7}{2} m/s \]