Question

Attached here is the question:

Answer 1

Answer:

t = 6 [s]

Explanation:

In order to solve this problem we must first use this equation of kinematics.

$v_{f}^{2}=v_{o}^{2} +2*a*x$

where:

Vf = final velocity = 0 (the car comes to rest)

Vo = initial velocity = 72 [km/h]

a = acceleration [m/s²]

x = distance = 60 [m]

First we must convert the velocity from kilometers per hour to meters per second.

$72 [\frac{km}{h}]*\frac{1000m}{1km} *\frac{1h}{3600s} =20 [m/s]$

$0=(20)^{2} -2*a*60\\400 = 120*a\\a=3.33[m/s^{2} ]$

Now using this other equation of kinematics.

$v_{f}=v_{o}-a*t$

0 = 20-3.33*t

t = 6[s]