Question

The velocity of an automobile starting from rest is given by the equation below, where v is measured in feet per second and t is the time in seconds. (Round your answers to three decimal places.)
v(t) = 105t/ 6t + 17

a. Find the acceleration at 5 seconds.
b. Find the acceleration at 10 seconds
c. Find the acceleration at 20 seconds

Answer 1

Answer:

a. At t = 5 s

$a(5)=\frac{1785}{\left(6(5)+17\right)^2}=\frac{1785}{2209}\approx0.808 \frac{ft}{s^2}$

b. At t = 10 s

$a(10)=\frac{1785}{\left(6(10)+17\right)^2}=\frac{255}{847}\approx0.301 \frac{ft}{s^2}$

c. At t = 20 s

$a(20)=\frac{1785}{\left(6(20)+17\right)^2}=\frac{1785}{18769}\approx0.095 \frac{ft}{s^2}$

Step-by-step explanation:

We know that the velocity function is given by

                                                   $v(t)=\frac{105t}{6t+17}$

Acceleration is the rate of change of velocity so we take the derivative of the velocity function with respect to time.

$a(t)=\frac{dv}{dt}=\frac{d}{dt} (\frac{105t}{6t+17})$

$\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'\\\\105\frac{d}{dt}\left(\frac{t}{6t+17}\right)\\\\\mathrm{Apply\:the\:Quotient\:Rule}:\quad \frac{d}{{dx}}\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \frac{{\frac{d}{{dx}}f\left( x \right)g\left( x \right) - f\left( x \right)\frac{d}{{dx}}g\left( x \right)}}{{g^2 \left( x \right)}}$

$105\cdot \frac{\frac{d}{dt}\left(t\right)\left(6t+17\right)-\frac{d}{dt}\left(6t+17\right)t}{\left(6t+17\right)^2}\\\\105\cdot \frac{1\cdot \left(6t+17\right)-6t}{\left(6t+17\right)^2}\\\\a(t)=\frac{1785}{\left(6t+17\right)^2}$

a. At t = 5 s

$a(5)=\frac{1785}{\left(6(5)+17\right)^2}=\frac{1785}{2209}\approx0.808 \frac{ft}{s^2}$

b. At t = 10 s

$a(10)=\frac{1785}{\left(6(10)+17\right)^2}=\frac{255}{847}\approx0.301 \frac{ft}{s^2}$

c. At t = 20 s

$a(20)=\frac{1785}{\left(6(20)+17\right)^2}=\frac{1785}{18769}\approx0.095 \frac{ft}{s^2}$