The driver of a car slams on the brakes, causing the car to slow down at a rate of

26ft/s2 as the car skids 256ft to a stop.

How long does the car take to stop?

What was the car's initial speed?

ft/s 4

A. The time taken for the car to stop is 3.14 secs

B. The initial velocity is 81.64 ft/s

Explanation:

Data obtained from the question include:

Acceleration (a) = 26ft/s2

Distance (s) = 256ft

Final velocity (V) = 0

Time (t) = ?

Initial velocity (U) = ?

A. Determination of the time taken for the car to stop.

Let us obtain an express for time (t)

Acceleration (a) = Velocity (V) / time (t)

a = V/t

Velocity (V) = distance (s) / time (t)

V = s/t

a = s/t^2

Cross multiply

a x t^2 = s

Divide both side by a

t^2 = s/a

Take the square root of both side

t = √ (s/a)

Now we can obtain the time as follow

Acceleration (a) = 26ft/s2

Distance (s) = 256ft

Time (t) = ... ?

t = √ (s/a)

t = √ (256/26)

t = 3.14 secs

Therefore, the time taken for the car to stop is 3.14 secs

B. Determination of the initial speed of the car.

V = U + at

Final velocity (V) = 0

Deceleration (a) = - 26ft/s2

Time (t) = 3.14 sec

Initial velocity (U) =.?

0 = U - 26x3.14

0 = U - 81.64

Collect like terms

U = 81.64 ft/s

Therefore, the initial velocity is 81.64 ft/s