Assertion (A): A particle moves in a straight line with constant acceleration. The average velocity of this particle can not be zero in any time interval.
Reason (R): For a particle moving in straight line, the average velocity in a time interval is always ((frac{u+v}{2})), where (u) and (v) are initial and final velocities of the particle in given time interval.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
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Solution: (A) is false; average velocity can be zero if displacement is zero (e.g., object returns to start with constant acceleration). (R) is false; the formula \(v_{avg} = \frac{u+v}{2}\) is only valid for constant acceleration, not 'always' for any straight line motion.
Assertion (A): At any instant, acceleration of a body can change its direction without any change in the direction of velocity.
Reason (R): At any instant, direction of acceleration is same as that of direction of change in velocity vector at that instant.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
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Concept: Relationship between acceleration and velocity.
Formula: \(vec{a} = \frac{d\vec{v}}{dt}\).
Solution: (A) is true. For example, a car moving straight can accelerate forward, then brake (accelerate backward) while its velocity direction remains forward. (R) is true; acceleration is defined as the rate of change of velocity, so its direction is the same as the direction of the change in velocity. (R) correctly explains (A) by providing the fundamental definition.
A particle covers half of its total distance with speed \( v_1 \) and the rest half distance with speed \( v_2 \). Its average speed during the complete journey is:
[2011 Mains]
1. \( \frac{v_1 v_2}{v_1 + v_2} \)
2. \( \frac{2v_1 v_2}{v_1 + v_2} \)
3. \( \frac{v_1 v_2^2}{v_1 + v_2^2} \)
4. \( \frac{v_1 v_2}{2} \)
View Answer
Concept: Average speed is total distance divided by total time. Let total distance be \( D \). Time for first half: \( t_1 = \frac{D/2}{v_1} \). Time for second half: \( t_2 = \frac{D/2}{v_2} \). Total time \( T = t_1 + t_2 = D \left( \frac{v_1 + v_2}{2v_1 v_2} \right) \). Average speed \( = \frac{D}{T} = \frac{2v_1 v_2}{v_1 + v_2} \).
A car moves from X to Y with a uniform speed \( v_u \) and returns to Y with a uniform speed \( v_d \). The average speed for this round trip is:
(2007)
1. \( \sqrt{v_u v_d} \)
2. \( \frac{v_d v_u}{v_d + v_u} \)
3. \( \frac{v_u + v_d}{2} \)
4. \( \frac{2v_d v_u}{v_d + v_u} \)
View Answer
Concept: Average speed is total distance divided by total time. Let distance from X to Y be \( D \). Time taken to go to Y: \( t_u = D/v_u \). Time taken to return to X: \( t_d = D/v_d \). Total distance \( = 2D \). Total time \( = t_u + t_d = D/v_u + D/v_d = D \frac{v_u + v_d}{v_u v_d} \). Average speed \( = \frac{2D}{D \frac{v_u + v_d}{v_u v_d}} = \frac{2v_u v_d}{v_u + v_d} \).
A bus travelling the first one-third distance at a speed of 10 km/h, the next one-third at 20 km/h and at last one-third at 60 km/h. The average speed of the bus is:
(1997)
1. 9 km/h
2. 16 km/h
3. 18 km/h
4. 48 km/h
View Answer
Concept: Average speed is total distance over total time. Let total distance be \( D \). The time taken for each one-third distance is \( t_1 = \frac{D/3}{10} = \frac{D}{30} \), \( t_2 = \frac{D/3}{20} = \frac{D}{60} \), \( t_3 = \frac{D/3}{60} = \frac{D}{180} \). Total time \( T = t_1+t_2+t_3 = D \left( \frac{6+3+1}{180} \right) = \frac{D}{18} \). Average speed \( = \frac{D}{T} = \frac{D}{D/18} = 18 \text{ km/h} \).
A car moves a distance of 200 m. It covers the first half of the distance at speed 40 km/h and the second half of distance at speed \( v \). The average speed is 48 km/h. The value of \( v \) is:
(1991)
1. 56 km/h
2. 60 km/h
3. 50 km/h
4. 48 km/h
View Answer
Concept: Average speed for two equal distance segments is \( v_{avg} = \frac{2v_1 v_2}{v_1 + v_2} \). Given \( v_{avg} = 48 \text{ km/h} \), \( v_1 = 40 \text{ km/h} \), and \( v_2 = v \). Substitute these values: \( 48 = \frac{2 \times 40 \times v}{40 + v} \). Simplify and solve for \( v \): \( 48(40+v) = 80v \) leads to \( 32v = 1920 \), so \( v = 60 \text{ km/h} \).
A particle is moving such that its position coordinates (x, y) are: \((2\text{ m}, 3\text{ m})\text{ at time } t = 0,\) \((6\text{ m}, 7\text{ m})\text{ at time } t = 2\text{ s}\) and \((13\text{ m}, 14\text{ m})\text{ at time } t = 5\text{ s}\). Average velocity \((\vec{V}_{av})\text{ from } t = 0\text { to } t = 5\text{ s}\) is:
(2014)
1. \(\frac{1}{5}(13\hat{i}+14\hat{j})\)
2. \(\frac{7}{3}(\hat{i}+\hat{j})\)
3. \(2(\hat{i}+\hat{j})\)
4. \(\frac{11}{5}(\hat{i}+\hat{j})\)
View Answer
Average velocity is \(\vec{V}_{av} = \frac{\Delta \vec{r}}{\Delta t} = \frac{\vec{r}_f - \vec{r}_i}{t_f - t_i}\). Initial position at \(t_i = 0\text{ s}\) is \(\vec{r}_i = 2\hat{i} + 3\hat{j}\). Final position at \(t_f = 5\text{ s}\) is \(\vec{r}_f = 13\hat{i} + 14\hat{j}\). So, \(vec{V}_{av} = \frac{(13\hat{i} + 14\hat{j}) - (2\hat{i} + 3\hat{j})}{5 - 0} = \frac{11\hat{i} + 11\hat{j}}{5} = \frac{11}{5}(\hat{i} + \hat{j})\) m/s.