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.
