The velocity versus time curve of a moving point is as given below. The maximum acceleration and displacement are respectively

Question

The velocity versus time curve of a moving point is as given below. The maximum acceleration and displacement are respectively (A) 1 cm / s 2, 1700 cm (B) 2 cm / s 2, 17 m (C) 3 cm / s 2, 1700 cm (D) 4 cm / s 2, 17 m

Tardigrade Admin · Accepted Answer

The maximum acceleration will occur in the duration 30 s to 40 s. So a =( v 2- v 1/ t 2- t 1)=(60-20/40-30)=4 cm / sec 2

Q. The velocity versus time curve of a moving point is as given below. The maximum acceleration and displacement are respectively

$1 c m / s^{2}, 1700 c m$

$2 c m / s^{2}, 17 m$

$3 c m / s^{2}, 1700 c m$

$4 c m / s^{2}, 17 m$

The maximum acceleration will occur in the duration $30 s$ to $40 s$ . So
$a = \frac{v _{2} - v _{1}}{t _{2} - t _{1}} = \frac{60 - 20}{40 - 30} = 4 c m / sec^{2}$

Q. The velocity versus time curve of a moving point is as given below. The maximum acceleration and displacement are respectively

1cm/s2,1700cm

2cm/s2,17m

3cm/s2,1700cm

4cm/s2,17m

The maximum acceleration will occur in the duration 30s to 40s. So a=t2​−t1​v2​−v1​​=40−3060−20​=4cm/sec2

$1 c m / s^{2}, 1700 c m$

$2 c m / s^{2}, 17 m$

$3 c m / s^{2}, 1700 c m$

$4 c m / s^{2}, 17 m$

The maximum acceleration will occur in the duration $30 s$ to $40 s$ . So
$a = \frac{v _{2} - v _{1}}{t _{2} - t _{1}} = \frac{60 - 20}{40 - 30} = 4 c m / sec^{2}$