Question

A particle moves in a straight line and has acceleration given by a(t) = 12t + 10. Its initial velocity is v(0) = −5 cm/s and its initial displacement is s(0) = 9 cm. Find its position function, s(t).

Answer 1

Answer:

The position function is $s_{t}=2t^3+5t^2-5t+9$.

Explanation:

Given that,

Acceleration $a =12t+10$

Initial velocity $v_{0} = -5\ cm/s$

Initial displacement $s_{0}=9\ cm$

We know that,

The acceleration is the rate of change of velocity of the particle.

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

The velocity is the rate of change of position of the particle

$v=\dfrac{dx}{dt}$

We need to calculate the the position

The acceleration is

$a_{t} = 12t+10$

$\dfrac{dv}{dt} = 12t+10$

$a_{t}=dv=(12t+10)dt$

On integration both side

$\int{dv}=\int{(12t+10)}dt$

$v_{t}=6t^2+10t+C$

At t = 0

$v_{0}=0+0+C$

$C=-5$

Now, On integration again both side

$v_{t}=\int{ds_{t}}=\int{(6t^2+10t-5)}dt$

$s_{t}=2t^{3}+5t^2-5t+C$

At t = 0

$s_{0}=0+0+0+C$

$C=9$

$s_{t}=2t^3+5t^2-5t+9$

Hence, The position function is $s_{t}=2t^3+5t^2-5t+9$.