Sure, let's solve this step by step.
### Problem Statement
Given:
1. Mass of the car, [tex]\( m = 1200 \, \text{kg} \)[/tex]
2. Initial speed of the car, [tex]\( v_0 = 50 \, \text{km/h} \)[/tex]
3. Stopping distance, [tex]\( d = 20 \, \text{m} \)[/tex]
4. The car comes to a stop, so the final speed [tex]\( v_f = 0 \, \text{m/s} \)[/tex]
We need to find:
a) The magnitude of the braking force.
b) The time required to stop.
### Step-by-Step Solution
Step 1: Convert the speed from km/h to m/s
[tex]\[ v_0 = 50 \, \text{km/h} \][/tex]
[tex]\[ v_0 = 50 \times \frac{1000 \, \text{m}}{1 \, \text{km}} \times \frac{1 \, \text{h}}{3600 \, \text{s}} \][/tex]
[tex]\[ v_0 = \frac{50 \times 1000}{3600} \][/tex]
[tex]\[ v_0 = \frac{50000}{3600} \][/tex]
[tex]\[ v_0 \approx 13.89 \, \text{m/s} \][/tex]
Step 2: Calculate the magnitude of the braking force
Using the work-energy principle:
The work done by the braking force [tex]\( W \)[/tex] is equal to the change in kinetic energy ([tex]\( \Delta KE \)[/tex]).
[tex]\[ \Delta KE = \text{Initial Kinetic Energy} - \text{Final Kinetic Energy} \][/tex]
[tex]\[ \Delta KE = \frac{1}{2} m v_0^2 - \frac{1}{2} m v_f^2 \][/tex]
Since [tex]\( v_f = 0 \)[/tex]:
[tex]\[ \Delta KE = \frac{1}{2} m v_0^2 \][/tex]
The work done by the braking force is also equal to the force multiplied by the distance:
[tex]\[ W = F_d \times d \][/tex]
Equating both expressions for work:
[tex]\[ F_d \times d = \frac{1}{2} m v_0^2 \][/tex]
Solving for the braking force [tex]\( F_d \)[/tex]:
[tex]\[ F_d = \frac{\frac{1}{2} m v_0^2}{d} \][/tex]
Substitute the known values:
[tex]\[ F_d = \frac{\frac{1}{2} \times 1200 \, \text{kg} \times (13.89 \, \text{m/s})^2}{20 \, \text{m}} \][/tex]
After evaluating the above expression, we get:
[tex]\[ F_d \approx 5787.04 \, \text{N} \][/tex]
So, the magnitude of the braking force is approximately [tex]\( 5787.04 \, \text{N} \)[/tex].
Step 3: Calculate the time required to stop
First, we need to find the acceleration [tex]\( a \)[/tex] using Newton's second law:
[tex]\[ F_d = m \times a \][/tex]
Solving for [tex]\( a \)[/tex]:
[tex]\[ a = \frac{F_d}{m} \][/tex]
[tex]\[ a = \frac{5787.04 \, \text{N}}{1200 \, \text{kg}} \][/tex]
[tex]\[ a \approx 4.82 \, \text{m/s}^2 \][/tex]
Next, use the first equation of motion to find the time [tex]\( t \)[/tex]:
[tex]\[ v_f = v_0 + a \times t \][/tex]
Since [tex]\( v_f = 0 \)[/tex]:
[tex]\[ 0 = 13.89 \, \text{m/s} - 4.82 \, \text{m/s}^2 \times t \][/tex]
[tex]\[ t = \frac{13.89 \, \text{m/s}}{4.82 \, \text{m/s}^2} \][/tex]
[tex]\[ t \approx 2.88 \, \text{s} \][/tex]
So, the time required to stop the car is approximately [tex]\( 2.88 \)[/tex] seconds.
### Final Answers
a) The magnitude of the braking force is approximately [tex]\( 5787.04 \, \text{N} \)[/tex].
b) The time required to stop the car is approximately [tex]\( 2.88 \)[/tex] seconds.
