Question

Given s left parenthesis t right parenthesis equals 5 t squared plus 5 ts(t)=5t2+5t​, find

Answer 1

Answer:

The velocity function is $v(t)=10t+5$.

The acceleration function is $a(t)=10$.

When t = 44​, the velocity is $v(44)=445 \:\frac{ft}{s}$.

When t = 44​, the acceleration is $a(44)=10\: \frac{ft}{s^2}$.

Step-by-step explanation:

We know that the position function is given by

$s(t)=5t^2+5t$

Velocity is defined as the rate of change of position or the rate of displacement. If you take the derivative of the position function you get the instantaneous velocity function.

$v(t)=\frac{ds}{dt}$

Acceleration is defined as the rate of change of velocity. If you take the derivative of the instantaneous velocity function you get the instantaneous acceleration function.

$a(t)=\frac{dv}{dt}$

The instantaneous velocity function is given by

$v(t)=\frac{d}{dt} s(t)=\frac{d}{dt}(5t^2+5t)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(5t^2\right)+\frac{d}{dt}\left(5t\right)\\\\\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}\\\\v(t)=10t+\frac{d}{dt}\left(5t\right)\\\\\mathrm{Apply\:the\:common\:derivative}:\quad \frac{d}{dt}\left(t\right)=1\\\\v(t)=10t+5$

The instantaneous acceleration function is given by

$a(t)=\frac{dv}{dt} =\frac{d}{dt}(10t+5)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\a(t)=\frac{d}{dt}\left(10t\right)+\frac{d}{dt}\left(5\right)\\\\a(t)=10$

To find the velocity and acceleration when t = 44, we substitute this value into the velocity and acceleration functions

$v(44)=10(44)+5\\v(44)=445 \:\frac{ft}{s}$

$a(44)=10\: \frac{ft}{s^2}$