Question

A 500kg car is traveling along a highway at 72.0km/he. If the driver immediately applies his brakes and the car comes to rear in a distance of 55.0 m, What is the average force acted on the car during de-acceleration?

Answer 1

First, let's calculate the average acceleration, using Torricelli's equation:

`V^2=V^2_0+2\cdot a\cdot d`

Where V is the final speed, V0 is the initial speed, a is the acceleration and d is the distance traveled.

So, for V = 0, V0 = 20 m/s (72 km/h divided by 3.6 to convert to m/s) and d = 55 m, we have:

`\begin{gathered} 0=20^2+2\cdot a\cdot55 \\ 110a=-400 \\ a=-3.636\text{ m/s}^2 \end{gathered}`

Now, to calculate the average force, let's use Newton's second law:

`\begin{gathered} F=m\cdot a \\ F=500\cdot3.636 \\ F=1818\text{ N} \end{gathered}`