The velocity versus time curve of a moving point is as shown in the figure. The maximum acceleration is

Question

The velocity versus time curve of a moving point is as shown in the figure. The maximum acceleration is (A) 1 m / s 2 (B) 4 m / s 2 (C) 2 m / s 2 (D) 1.5 m / s 2

Tardigrade Admin · Accepted Answer

Maximum acceleration is represented by maximum slope of velocity-time graph. It is for portion C D of graph. Slope of C D=(60-20/40-30)=(40/10)=4 m / s 2

Q. The velocity versus time curve of a moving point is as shown in the figure. The maximum acceleration is

$1 m / s^{2}$

$4 m / s^{2}$

$2 m / s^{2}$

$1.5 m / s^{2}$

Maximum acceleration is represented by maximum slope of velocity-time graph. It is for portion $C D$ of graph.
Slope of $C D = \frac{60 - 20}{40 - 30} = \frac{40}{10} = 4 m / s^{2}$

Q. The velocity versus time curve of a moving point is as shown in the figure. The maximum acceleration is

1m/s2

4m/s2

2m/s2

1.5m/s2

Maximum acceleration is represented by maximum slope of velocity-time graph. It is for portion CD of graph. Slope of CD=40−3060−20​=1040​=4m/s2

$1 m / s^{2}$

$4 m / s^{2}$

$2 m / s^{2}$

$1.5 m / s^{2}$

Maximum acceleration is represented by maximum slope of velocity-time graph. It is for portion $C D$ of graph.
Slope of $C D = \frac{60 - 20}{40 - 30} = \frac{40}{10} = 4 m / s^{2}$