JEE MainPhysicsMotion in One Dimension
A particle starts from the origin at t=0 . Its velocity-time ( v-t ) graph consists of four continuous segments of equal time intervals t₀ : (1) From t=0 to t=t₀ , velocity increases linearly from zero to v₀ . (2) From t=t₀ to t=2t₀ , velocity remains constant at v₀ . (3) From t=2t₀ to t=3t₀ , velocity decreases linearly from v₀ to zero. (4) From t=3t₀ to t=4t₀ , velocity continues to decrease linearly from zero to -
Options
- AFrom t=0 to t₀ , x increases with upward concavity; from t₀ to 2t₀ , x increases linearly; from 2t₀ to 3t₀ , x
- BFrom t=0 to t₀ , x increases linearly; from t₀ to 2t₀ , x remains constant; from 2t₀ to 3t₀ , x decreases line
- CFrom t=0 to t₀ , x increases with upward concavity; from t₀ to 2t₀ , x increases linearly; from 2t₀ to 3t₀ , x
- DFrom t=0 to t₀ , x increases with downward concavity; from t₀ to 2t₀ , x increases linearly; from 2t₀ to 3t₀ ,
Correct answer
A. From t=0 to t₀ , x increases with upward concavity; from t₀ to 2t₀ , x increases linearly; from 2t₀ to 3t₀ , x
Step-by-step solution
The slope of the position-time ( x-t ) graph represents velocity ( v = dx dt ). For t=0 to t₀ : v > 0 and increasing. Thus, x increases and the slope of the x-t graph increases, which means upward concavity. For t=t₀ to 2t₀ : v = v₀ > 0 (constant). Thus, x increases linearly with a constant positive slope. For t=2t₀ to 3t₀ : v > 0 but decreasing to zero. Thus, x continues to increase, but the slope decreases, meaning downward concavity. At t=3t₀ , v=0 , so x reaches a local maximum. For t=3t₀ to 4t₀ : v Answer: Fro