Question

You are traveling on an interstate highway at the posted speed limit of 70 mph when you see that the traffic in front of you has stopped due to an accident up ahead. You step on your brakes to slow down as quickly as possible. Assume that you to slow down to 30 mph in about 5 seconds. A) With this same average acceleration, how much longer would it take you to stop?

B) What total distance would you travel from when you first apply the brakes until the car stops?

Answer 1

With the same average acceleration, it would take approximately 4.97 seconds longer for you to stop. You would travel a total distance of 78.2 meters from when you first apply the brakes until the car stops.

Step-by-step explanation:

To determine how much longer it would take for you to stop with the same average acceleration of 3.6 m/s², we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

If you initially step on your brakes at 70 mph (31.3 m/s) and slow down to 30 mph (13.4 m/s) in 5 seconds, the change in velocity (Δv) would be 31.3 m/s - 13.4 m/s = 17.9 m/s. To calculate the time it would take to stop with the same acceleration, we can rearrange the equation to t = (v - u) / a. Plugging in the values, t = (0 - 17.9 m/s) / -3.6 m/s² = 4.97 seconds.

Therefore, it would take approximately 4.97 seconds longer for you to stop at the same acceleration.

To calculate the total distance you would travel from when you first apply the brakes until the car stops, we can use the equation s = ut + 0.5at², where s is the distance, u is the initial velocity, t is the time, and a is the acceleration.

Plugging in the values, s = 31.3 m/s * 5s + 0.5 * -3.6 m/s² * (5s)² = 78.2 m. Therefore, you would travel a total distance of 78.2 meters from when you first apply the brakes until the car stops.

Answer 2

Answer:8.75 s,

136.89 m

Step-by-step explanation:

Given

Initial velocity
`=70 mph\approx 31.29 m/s`

velocity after 5 s is
`30 mph\approx 13.41 m/s`

Therefore acceleration during these 5 s

`a=(v-u)/(t)`

`a=(13.41-31.29)/(5)=-3.576 m/s^2`

therefore time required to stop

v=u+at

here v=final velocity =0 m/s

initial velocity =31.29 m/s

`0=31.29-3.576* t`

`t=(31.29)/(3.576)=8.75 s`

(b)total distance traveled before stoppage

`v^2-u^2=2as`

`0^2-31.29^2=2* (-3.576)\cdot s`

s=136.89 m