Question

A little help please?

Spotting a police car, you brake your Porsche from a speed of 100 km/h
to a speed of 80.0 km/h during a displacement of 88.0 m, at a constant acceleration.

a. What is that acceleration?
b. How much time is required for the given decrease in speed?

Answer 1

a. `a = -1.579 m/s^2`
b. `t = 3.52 s`

Explanation:

We have velocity in km/hr and displacement in m. Since we have to make a unit conversion, might as well do it so it will yield the answer in the normal (SI) units for acceleration.

`100 (km)/(hr) * (1000 m)/(1 km)*(1 hr)/(3600 s) = 27.78 m/s`

`80.0 (km)/(hr) * (1000 m)/(1 km)*(1 hr)/(3600 s) = 22.22 m/s`

a. Using the formula of motion `v^2 = u^2 + 2*a*d`

#(22.22 m/s)^2 = (27.78 m/s)^2 + 2a88.0 m

`493.7 m^2/s^2 = 771.7 m^2/s^2 + 176 m*a`

`a = (493.7 m^2/s^2 - 771.7 m^2/s^2)/(176 m) = -1.579 m/s^2`

b. Since the acceleration was constant, we can use the average velocity, and the distance to determine the time.

`t = (88.0 m)/(1/2*(27.78 m/s + 22.22 m/s)) = 3.52 s`

As a double-check, if we convert `v = u + a*t` to `t = (v - u)/a` and plug in the data, let's see if we get the same time.

`t = (22.22 m/s - 27.78 m/s^2)/(-1.579 m/s^2) = 3.52 s`

I hope this helps,
Steve