The driver of a car slams on the brakes, causing the car to
slow down at a rate of 17ft/s2 as the car skids 175ft to a
stop.
How long does the car take to stop?

What was the car's initial speed?

After the driver of a car slams on the brakes, which causes the car to stop after traveling 175 ft with a slow down rate of 17 ft/s², we have:

a) The car stops after 4.54 s.

b) The car's initial speed is 77.14 ft/s.

We can calculate the time at which the car stops with the following kinematic equation:

  (1)

Where:                

t: is the time =?

: is the initial velocity

: is the final velocity = 0 (the car stops)

a: is the acceleration = -17 ft/s² (the minus sign is because the car is slowing down its speed)    

First, we need to find the initial velocity.

Where:    

d: is the distance traveled = 175 ft

Then, the initial velocity is:

Hence, the initial velocity is 77.14 ft/s.

Now, the time at which the car stops is (eq 1):

Therefore, the car stops after 4.54 s.

Find more about acceleration here:

I hope it helps you!

It takes 4.58 seconds for the car to stop and the initial speed of the car is 77.14 ft/s

Explanation:

Given:

Acceleration, a = -17 ft/s²

Distance, s  = 175 ft

Final velocity, v = 0

Time, t = ?

Initial speed, u = ?

We know:

v² - u² = 2as

(0)² - (u)² = 2 X - 17 X 175

u² = 5950

u = 77.14 ft/s

To calculate time:

v = u + at

0 = 77.14 - 17 X t

17t = 77.14

t = 4.58 sec

Therefore, It takes 4.58 seconds for the car to stop and the initial speed of the car is 77.14 ft/s

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west