A car with a mass of [tex]10^3[/tex] kilograms traveling at a speed of 50 kilometers/hour (13.9 meters/second) collides with a stationary car. The duration of the collision is 1 second. After the collision, the speed of the moving car is reduced to 5 kilometers/hour (1.4 meters/second). What is the value of the force involved in the collision?

A. [tex]1.2 \times 10^4[/tex] newtons
B. [tex]1.4 \times 10^4[/tex] newtons
C. [tex]1.6 \times 10^3[/tex] newtons
D. [tex]1.6 \times 10^2[/tex] newtons



Answer :

Alright, let's break this problem down step-by-step to find the force involved in the collision.

First, let’s recollect the given information:

- Mass of the car, [tex]\( m \)[/tex]: [tex]\( 10^3 \)[/tex] kilograms.
- Initial speed of the car, [tex]\( u \)[/tex]: 13.9 meters/second.
- Final speed of the car, [tex]\( v \)[/tex]: 1.4 meters/second.
- Duration of the collision, [tex]\( t \)[/tex]: 1 second.

We need to find the force involved in the collision. According to Newton's second law of motion, force [tex]\( F \)[/tex] is given by:
[tex]\[ F = m \times a \][/tex]
where [tex]\( a \)[/tex] is the acceleration (or deceleration in this case) of the car.

### Step 1: Calculate the Change in Velocity
The change in velocity ([tex]\( \Delta v \)[/tex]) is the difference between the final speed and the initial speed:
[tex]\[ \Delta v = v - u \][/tex]
[tex]\[ \Delta v = 1.4 \, \text{m/s} - 13.9 \, \text{m/s} \][/tex]
[tex]\[ \Delta v = -12.5 \, \text{m/s} \][/tex]

The negative sign indicates a decrease in speed.

### Step 2: Calculate the Acceleration
Acceleration ([tex]\( a \)[/tex]) is the change in velocity ([tex]\( \Delta v \)[/tex]) divided by the time ([tex]\( t \)[/tex]):
[tex]\[ a = \frac{\Delta v}{t} \][/tex]
[tex]\[ a = \frac{-12.5 \, \text{m/s}}{1 \, \text{s}} \][/tex]
[tex]\[ a = -12.5 \, \text{m/s}^2 \][/tex]

The negative sign indicates deceleration.

### Step 3: Calculate the Force
Using the mass ([tex]\( m \)[/tex]) and the acceleration ([tex]\( a \)[/tex]), we can calculate the force ([tex]\( F \)[/tex]):
[tex]\[ F = m \times a \][/tex]
[tex]\[ F = 10^3 \, \text{kg} \times (-12.5 \, \text{m/s}^2) \][/tex]
[tex]\[ F = -12500 \, \text{N} \][/tex]

The negative sign here means the force is in the direction opposite to the motion of the car.

### Conclusion
Since we are asked for the magnitude of the force:
[tex]\[ |F| = 12500 \, \text{N} \][/tex]

Comparing this to the given options, the correct answer is:
A. [tex]\(1.2 \times 10^4\)[/tex] newtons.