Question

A particle moves with acceleration function a(t) = 5 + 4t - 2t^2. Its initial velocity v(0) = 3 m/s and its initial displacement is s(0) = 10 m. Find its position after 4 seconds.

Answer 1

Answer:

Its position after 4 seconds is 62 meters.

Explanation:

It is given that,

The acceleration of the particle is given by equation :

$a(t)=5+4t-2t^2$

Also, $a=\dfrac{dv}{dt}$

$v=\int\limits {a.dt}$

$v=\int\limits {(5+4t-2t^2).dt}$

$v=5t+2t^2-\dfrac{2}{3}t^3+c$

At t = 0, $v(0)=3\ m/s$. So, c = 3

$v=5t+2t^2-\dfrac{2}{3}t^3+3$

Also, $v=\dfrac{ds}{dt}$, s is the position

$s=\int\limits {v.dt}$

$s=\int\limits {(5t+2t^2-\dfrac{2}{3}t^3+3).dt}$

$s=\dfrac{5}{2}t^2+\dfrac{2}{3}t^3-\dfrac{t^4}{6}+3t+c'$

At t = 0, $s(0)=10\ m$. So, c' = 10

$s=\dfrac{5}{2}t^2+\dfrac{2}{3}t^3-\dfrac{t^4}{6}+3t+10$

At t = 4 s

$s=\dfrac{5}{2}(4)^2+\dfrac{2}{3}(4)^3-\dfrac{(4)^4}{6}+3(4)+10$

s = 62 m

So, at t = 4 seconds the position of the particle is 62 meters. Hence, this is the required solution.