a 1500 kg car moving with a speed of 15 m/s collides with a utility pole and is brought to rest in 0.3 s.
Find the average force exerted on the car during the collision.
To find the average force exerted on the car during the collision, we can use Newton's second law of motion, which states that force (F) is equal to the mass (m) of an object multiplied by its acceleration (a).
However, since the car starts with a certain velocity and comes to rest, we need to calculate the deceleration first.
The initial velocity (u) of the car is 15 m/s, and the final velocity (v) is 0 m/s. The time (t) taken to come to rest is 0.3 seconds.
Using the equation of motion v = u + at, we can rearrange it to find the acceleration (a):
a = (v - u) / t
a = (0 - 15) m/s / 0.3 s
a = -50 m/s^2
The acceleration is negative because it represents deceleration (opposite direction to the initial motion).
Now we can calculate the average force (F) using Newton's second law:
F = m * a
F = 1500 kg * (-50 m/s^2)
F = -75,000 N
The negative sign indicates that the force is in the opposite direction to the car's initial motion, which makes sense because it brings the car to rest.
Therefore, the average force exerted on the car during the collision is 75,000 Newtons (N) and acts in the opposite direction to the car's initial motion.