Question

A car traveling 56 km/h is 22.0 m from a barrier when the driver slams on the brakes. The car hits the barrier 2.03 s later.

(a) What was the car's constant deceleration before impact?
m/s²
(b) How fast was the car traveling at impact?
m/s

Answer 1

First, you need to pass km/h to m/s:

56 km/h x 1000 m/km x 1h/3600 s = 15.56 m/s

Now, let's take all the data, calling "x" distance, and Vi = innitial velocity, Vf = final velocity.

a) to get acceleration, use the following equation:

x = Vixt + at²/2 ----> Then solve for a:

22 = 15.56 x 2.03 + (2.03)² a /2

22 - 31.59 = 4.1209 a / 2

-9.59 x 2 / 4.1209 = a

a = -4.65 m/s²

b) Now with the value of acceleration, the Vf can be calculated with the following equations:

Vf = Vi + at -----or Vf² = Vi² + 2ax

You may use whenever you like the most. In this case, I'll use the first one:

Vf = 15.56 - 4.65x2.03

Vf = 6.1205 m/s

If you use the second equation, you should get the same result (or similar):

Vf = [(15.56)² - (2x4.65x22)]^1/2

Vf = (242.1136 - 204.6)^1/2 = 37.5136^1/2

Vf = 6.1248 m/s

You can actually see, that both results are correct and very similar.

Answer 2

First, you need to pass km/h to m/s:

56 km/h x 1000 m/km x 1h/3600 s = 15.56 m/s

Now, let's take all the data, calling "x" distance, and Vi = innitial velocity, Vf = final velocity.

a) to get acceleration, use the following equation:

x = Vixt + at2/2 ----> Then solve for a:

22 = 15.56 x 2.03 + (2.03)2 a /2

22 - 31.59 = 4.1209 a / 2

-9.59 x 2 / 4.1209 = a

a = -4.65 m/s2

b) Now with the value of acceleration, the Vf can be calculated with the following equations:

Vf = Vi + at -----or Vf2 = Vi2 + 2ax

You may use whenever you like the most. In this case, I'll use the first one:

Vf = 15.56 - 4.65x2.03

Vf = 6.1205 m/s

If you use the second equation, you should get the same result (or similar):

Vf = [(15.56)2 - (2x4.65x22)]1/2

Vf = (242.1136 - 204.6)1/2 = 37.51361/2

Vf = 6.1248 m/s