A car is traveling at [tex][tex]$20 \, \text{m/s}$[/tex][/tex]. It makes an emergency stop to avoid hitting a badger in the road. If the car has a mass of [tex][tex]$1000 \, \text{kg}$[/tex][/tex] and it travels [tex][tex]$48 \, \text{m}$[/tex][/tex] before it comes to a stop, how much braking force did the driver apply? Use the equations below to help you and give your answer to the nearest hundred.

Enter your answer as a number in [tex]N[/tex].

[tex]v^2 - u^2 = 2ad[/tex]



Answer :

Certainly! Let's solve the problem step-by-step.

1. Identify the given values:
- Initial velocity ([tex]\(u\)[/tex]) = 20 m/s
- Final velocity ([tex]\(v\)[/tex]) = 0 m/s (the car comes to a stop)
- Distance ([tex]\(d\)[/tex]) = 48 m
- Mass ([tex]\(m\)[/tex]) = 1000 kg

2. Determine the acceleration ([tex]\(a\)[/tex]) using the kinematic equation:
[tex]\[ v^2 - u^2 = 2 a d \][/tex]

Rearrange the equation to solve for acceleration ([tex]\(a\)[/tex]):
[tex]\[ a = \frac{v^2 - u^2}{2 d} \][/tex]

Substitute the given values into the equation:
[tex]\[ a = \frac{(0)^2 - (20)^2}{2 \times 48} \][/tex]
[tex]\[ a = \frac{-400}{96} \][/tex]
[tex]\[ a = -4.166666666666667 \, \text{m/s}^2 \][/tex]

The negative sign indicates that the acceleration is actually deceleration (i.e., the car is slowing down).

3. Calculate the braking force ([tex]\(F\)[/tex]) using Newton's second law:
[tex]\[ F = m \cdot a \][/tex]
Substitute the mass and the calculated acceleration into the equation:
[tex]\[ F = 1000 \, \text{kg} \times (-4.166666666666667 \, \text{m/s}^2) \][/tex]
[tex]\[ F = -4166.666666666667 \, \text{N} \][/tex]

The negative sign in the force indicates that it is a braking force (opposite to the direction of motion).

4. Round to the nearest hundred:
[tex]\[ F \approx -4200 \, \text{N} \][/tex]

Therefore, the braking force applied by the driver is approximately [tex]\( -4200 \)[/tex] N.