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JEE Main 2024 (6 Apr, Shift 1) · Distance, Displacement & Average Speed

Distance, Displacement & Average Speed question from JEE Main 2024 (6 Apr, Shift 1)

Q8JEE Main 2024 · 6 APR · SHIFT 1Medium

Distance, Displacement & Average Speed

A train starting from rest first accelerates uniformly up to a speed of $80\text{ km/h}$ for time $t$ , then it moves with a constant speed for time $3t$ . The average speed of the train for this duration of journey will be (in $\text{km/h}$ ):

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Answer: (a)

Accelerating from rest to 80 km/h in time $t$ : distance $= \frac{0 + 80}{2}t = 40t$ .
Constant speed for $3t$ : distance $= 80 \times 3t = 240t$ .
$v_{avg} = \frac{40t + 240t}{4t} = 70$ km/h
So the answer is (a) 70.

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