# Answer in Calculus for Delmundo #209118

An MBDA official investigated an accident in Roman highway. They saw a 200-

feet-long skid mark which was produced from the time driver stepped on the car

break to the time it fully stopped. During the investigation, the driver told him that

after stepping on the car break, it decelerated at a speed of 16 / 2

. The official

wanted to know how fast the car was before the accident happened. Determine the

car’s velocity before the break is applied to help them dig on what caused the

accident.

t= 0 t= n

vi=? vf= 0

Si= 0 Sf= 200ft

First, we have to consider and solve the equation that involves the constant deacceleration as the change on the speed velocity which is:

“a=frac{dv}{dt} implies int_{v_{i}}^{v} dv=aint_{t=0}^{t} dtimplies v=at+v_{i}”

“v_{f}-v_{i}=a(n-0) implies v_{i}=-an=constant”

Since we also know that speed is related to the distance **s** we’ll go ahead and solve the next integral to find the time for this incident that occurred on a highway:

“v=frac{ds}{dt} implies int_{s_{i}=0}^{s_{f}}ds=int_{t=0}^{t=n}vdt=int_{t=0}^{t=n}(at+v_{i})dt”

“implies large{[s]_{s_{i}}^{s_{f}}=frac{a}{2}[t^2]_{t=0}^{t=n}}+ v_{i}[t]_{t=0}^{t=n}”

“implies s_{f}-0 = frac{a}{2}(n^2-0)+ (-an)(n-0)”

“implies s_{f} = frac{an^2}{2}+ -an^2= -frac{an^2}{2} implies n=sqrt{frac {-2s_{f}}{a}}”

After we substitute the distance S_{f}=200 m and the acceleration as a=-16 ft/s^{2} we find:

“n=sqrt{frac {-2s_{f}}{a}}=sqrt{frac {-2(200,ft)}{-16,largefrac{ft}{s^2}}}=sqrt{25,s^2}=5,s”

**In conclusion, the time that took for the vehicle to deaccelerate and stop was t = 5 s**.

Reference:

- Serway, R. A., & Jewett, J. W. (2018).
*Physics for scientists and engineers*. Cengage learning.