Answer to Question #135227 in Calculus for Aditi Yadav

Question

The equation of motion of a particle is s =((t^3)-3t) where s is in metres and t is in
seconds. Find
i) the velocity and acceleration as functions of t,
ii) the acceleration after 2 seconds,
iii) the acceleration, when the velocity is 0.

Answer 1

Consider the equation of motion of a particle $s=t^3-3t$

i)

Differentiate $s$ with respect to $t$ to obtain velocity function as,

$v=\frac{ds}{dt}$

=$\frac{d}{dt}(t^3-3t)$

$=\frac{d}{dt}(t^3)-3\frac{d}{dt}(t)$

$=3t^2-3(1)$

$=3t^2-3$

Therefore, the velocity function is $v=3t^2-3$

Now, differentiate velocity function with respect to $t$ to obtain acceleration function as,

$a=\frac{dv}{dt}$

$=\frac{d}{dt}(3t^2-3)$

$=3\frac{d}{dt}(t^2)-\frac{d}{dt}(3)$

$=3(2t)-0$

$=6t$

Therefore, the acceleration function is $a=6t$

ii)

Plug $t=2$ into acceleration function to obtain acceleration after $2$ seconds as,

$a=6(2)$

$=12$

Therefore, the acceleration after $2$ seconds is $a=12$ $m/s^2$

iii)

When velocity is , then find the time as,

$3t^2-3=0$

$3(t^2-1)=0$

$t^2-1=0$

$t=1$

Here, negative value of $t$ is neglected.

Now, plug $t=1$ into acceleration function to obtain acceleration after $1$ seconds as,

$a=6(1)$

$=6$

Therefore, the acceleration after $1$ seconds is $a=6$ $m/s^2$