Practicing Success
Practicing Success
CUET Preparation Today
The velocity versus time curve of a moving point is as given below. The maximum acceleration is
|
$1 cm/s^2$ $2 cm/s^2$ $3 cm/s^2$ $4 cm/s^2$ |
$4 cm/s^2$ |
The correct answer is Option (4) → $4 cm/s^2$ The maximum acceleration will occur in the duration 30s to 40s. So $a=\frac{v_2-v_1}{t_2-t_1}=\frac{60-20}{40-30}=4 cm/s^2$ |
