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/s²).

Solution

Physics sir · 4 steps

Fix one sign convention first: then the bullet's velocity starts at −100, grows to −200, and drops to zero.

1. Choose signs. Take upward as positive. The bullet moves down, so u = -100 m s⁻¹. Gravity also acts down, so a = -10 m s⁻².

   v = u + at = -100 - 10t

2. First 10 seconds. Velocity and acceleration point the same way, so the bullet speeds up. The graph is a straight line with slope -10, going from -100 at t = 0 to -100 - 100 = -200 m s⁻¹ at t = 10 s.
3. After hitting ground. The collision is perfectly inelastic, so the bullet stops at once. The velocity jumps vertically from -200 to 0 at t = 10 s and stays at 0 until t = 20 s.
4. So. Only graph (a) shows a line falling from -100 to -200 m s⁻¹, then a vertical jump to zero and a flat line up to 20 s.

The idea to keep. When velocity and acceleration point the same way, speed grows. On a v–t graph that means moving away from the time axis, whatever sign you chose.

Answer from the NTA final answer key. Solution by PrepWiser.