A car starts from Bengaluru, goes 50 km in a straight line towards south, immediately turns around and returns to Bengaluru. The time taken for this round trip is 2 hours. The magnitude of the average velocity of the car for this round trip
- (a)is 0.
- (b)is 50 km/hr.
- (c)is 25 km/hr.
- (d)cannot be calculated without knowing acceleration.
Answer
Why
Correct - A, the average velocity is 0. Average velocity is total DISPLACEMENT divided by time. The car leaves Bengaluru, goes 50 km south and returns to the very same starting point, so its net displacement over the round trip is zero. Zero displacement divided by 2 hours gives an average velocity of magnitude 0, however fast it actually moved. (Its average SPEED, using the total distance of 100 km, is 50 km/hr - a different quantity.)
Why the others are wrong
- (b)is 50 km/hr. — 50 km/hr is the average SPEED (total path length 100 km divided by 2 hours), not the average velocity, which uses displacement and is zero here.
- (c)is 25 km/hr. — 25 km/hr is neither the speed nor the velocity here; it looks like 50 km divided by 2 hours (one-way distance over round-trip time), a mismatched calculation.
- (d)cannot be calculated without knowing acceleration. — Displacement (0) and time (2 hours) are all that average velocity needs; the acceleration is irrelevant to it.
Concept
Velocity is a vector defined by displacement - the straight-line change in position - while speed is a scalar defined by the actual distance travelled along the path. For any journey that ends where it began, displacement, and hence average velocity, is exactly zero, even though the average speed is not.
The trap is to compute distance over time (a speed) and call it velocity. Ask 'where did it end up compared with where it started?' - the same point means zero displacement and zero average velocity.
Key facts
- Average velocity equals displacement divided by time, and is a vector.
- Average speed equals total distance divided by time, and is a scalar.
- Displacement for any closed round trip is zero.
- Here distance is 100 km and displacement is 0, so average speed is 50 km/hr but average velocity is 0.
Study next
Common traps
- Reporting distance over time as 'velocity' when the motion is not in one straight line.
- Assuming average speed also becomes zero for a round trip - only average velocity does.
Asked through a there-and-back or closed-loop journey, contrasting average velocity (zero) with average speed.
Related PYQs
Starting from rest a vehicle accelerates at the rate of 2 m/s² towards east for 10 s. It then stops suddenly. It then accelerates again at a rate of 4√2 m/s² for next 10 s towards south and then again comes to rest. The net displacement of the vehicle from the starting point is
- (a) 100 m
- (b) 200 m
- (c) 300 m
- (d) 400 m
Answer(c) 300 m
Same displacement-as-a-vector idea. There you add motions in different directions to get net displacement; here the round trip returns to the start, so displacement — and thus average velocity — is zero.
Practice
- practice — not a real PYQ
A boy runs once around a circular track of circumference 400 m in 100 s and returns to the start. His average velocity is
- (a)4 m/s
- (b)0
- (c)2 m/s
- (d)8 m/s
Answer(b) 0 - displacement over one complete loop is zero. - practice — not a real PYQ
For a moving body, the magnitude of average velocity equals the average speed only when
- (a)the body returns to its start
- (b)the motion is along a straight line without reversing
- (c)the acceleration is zero
- (d)the time taken is very small
Answer(b) the motion is along a straight line without reversing - then distance equals the magnitude of displacement.