JEE Main 2022 · 27 Jul · Shift 1 · Physics Q32

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A bullet is shot vertically downwards with an initial velocity of 100 m/s from a certain height. Within 10 s, the bullet reaches the ground and instantaneously comes to rest due to the perfectly inelastic collision. The velocity-time curve for total time t = 20 s will be:

(Take g = 10 m/s $^2$ ).

v falls linearly from −100 m/s to −200 m/s over 10 s, then is zero up to 20 s

v against t: a straight line from −100 m/s at t = 0 down to −200 m/s at 10 s, a jump to zero, then v = 0 up to 20 s.

v falls linearly from +100 m/s to 0 over 10 s, then is zero up to 20 s

v against t: a straight line from +100 m/s at t = 0 down to zero at 10 s, then v = 0 up to 20 s.

v falls linearly from 0 to −100 m/s over 10 s, then is zero up to 20 s

v against t: a straight line from zero at t = 0 down to −100 m/s at 10 s, a jump to zero, then v = 0 up to 20 s.

v rises linearly from −100 m/s to 0 over 10 s, then is zero up to 20 s

v against t: a straight line from −100 m/s at t = 0 up to zero at 10 s, then v = 0 up to 20 s.

Answer and worked solution

Official answer: A

Upward positive. $v=-100-10t$ falls from $-100$ to $-200$ m/s by 10 s, then the bullet stops: $v=0$ to 20 s.

(A)

Topics:Motion graphs Motion in a straight line

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JEE Main 2022 · 27 Jul · Shift 1, Physics question 32