Question

A body moves a distance of s=2t^3-4t+5 in t seconds.
a) what is the velocity of the body after 3 seconds ?
b)what is the acceleration after 4 seconds ?
c)what is the time when the velocity is zero?
d)what is the time when the acceleration is zero?

Answer 1

a) The instantaneous velocity of the body is the derivative of s(t) with respect to t:

$$v\left( t \right)=\frac{d}{dt}s\left( t \right)=\frac{d}{dt}\left( 2{{t}^{3}}-4t+5 \right)$$

$$=2\cdot 3{{t}^{2}}-4\cdot 1+0=6{{t}^{2}}-4$$

To find the velocity of the body after 3 seconds we substitute t=3 s:

$$v\left( 3 \right)=6\cdot {{3}^{2}}-4=50\,m/s$$

So the velocity of the body after 3 seconds is $v\left( 3 \right)=50\,m/s$

b) The instantaneous acceleration of the body is the derivative of v(t) with respect to t:

$$a\left( t \right)=\frac{d}{dt}v\left( t \right)=\frac{d}{dt}\left( 6{{t}^{2}}-4 \right)=6\cdot 2t+0=12t$$

To find the acceleration of the body after 4 seconds we substitute t=4 s:

$$a\left( 4 \right)=12\cdot 4=48\,m/{{s}^{2}}$$

So the acceleration after 4 seconds is $a\left( 4 \right)=48\,m/{{s}^{2}}$

c) To find the time when the velocity is zero, we equate v(t) to zero:

$$v\left( t \right)=6{{t}^{2}}-4=0$$

and solve this equation for t

$$6{{t}^{2}}=4$$

$${{t}^{2}}=\frac{4}{6}=\frac{2}{3}$$

$$t=\sqrt{\frac{2}{3}}\approx 0.82\,s$$

So the time when the velocity is zero is $t\approx 0.82\text{ }s$ .

d) To find the time when the acceleration is zero, we equate a(t) to zero:

$$a\left( t \right)=12t=0$$

and solve this equation for t: the solution is t=0 s.

So the time when the acceleration is zero is t=0 s.