A 862-kg car starts from rest on a horizontal roadway and accelerates eastward for 5.00 s when it reaches a speed of 30.0 m/s. What is the average force exerted on the car during this time?
To find the average force exerted on the car, we can use Newton's second law of motion, which states that the force (F) acting on an object is equal to the mass (m) of the object multiplied by its acceleration (a):
F = m * a
In this case, the car starts from rest and accelerates eastward, so the initial velocity (u) is 0 m/s and the final velocity (v) is 30.0 m/s. The time taken (t) is 5.00 seconds.
To find the acceleration, we can use the equation:
a = (v - u) / t
Plugging in the values:
a = (30.0 m/s - 0 m/s) / 5.00 s = 6.00 m/s^2
Now, we can calculate the force:
F = (862 kg) * (6.00 m/s^2) = 5172 N
Therefore, the average force exerted on the car during this time is 5172 Newtons.
To find the average force exerted on the car, we can use Newton's second law of motion, which states that the force exerted on an object is equal to the mass of the object multiplied by its acceleration.
In this case, the car's mass is given as 862 kg, and we need to find its acceleration. We can use the equation for acceleration:
acceleration = (final velocity - initial velocity) / time
Given that the car starts from rest, its initial velocity is 0 m/s. The final velocity is given as 30.0 m/s. The time taken to reach this velocity is given as 5.00 s.
Substituting these values into the equation, we get:
acceleration = (30.0 m/s - 0 m/s) / 5.00 s = 6.0 m/s^2
Now that we know the car's acceleration, we can use Newton's second law to find the average force:
force = mass * acceleration = 862 kg * 6.0 m/s^2 = 5,172 N
Therefore, the average force exerted on the car during the 5.00 s is 5,172 Newtons.