Question

The brochure advertising a sports car states that the car can be moving at 100.0 km/h, and stop in 37.19 meters. What is its average acceleration during a stop from that velocity? Express your answer in m/s². Consider the car's initial velocity to be a positive quantity

Answer 1

The average acceleration of the sports car during a stop is -0.75 m/s².

Step-by-step explanation:

To find the average acceleration of the sports car during a stop, we need to use the formula:
average acceleration = change in velocity / time taken

The car's initial velocity is 100.0 km/h and it stops in 37.19 meters. We need to convert the velocity to m/s by dividing it by 3.6. So, the initial velocity is 27.8 m/s. The final velocity is 0 m/s because the car comes to a stop.

Using the formula:
average acceleration = (final velocity - initial velocity) / time

We plug in the values:
average acceleration = (0 m/s - 27.8 m/s) / (37.19 m / 27.8 m/s) = -0.75 m/s²

Answer 2

The average acceleration of a sports car stopping from 100.0 km/h over 37.19 meters is calculated using kinematic equations and is found to be -10.37 m/s².

Step-by-step explanation:

To calculate the average acceleration of a sports car during a stop from a velocity of 100.0 km/h to 0 km/h over a distance of 37.19 meters, we need to use the kinematic equations of motion. The velocity needs to be converted from km/h to m/s.

Firstly, convert the initial velocity (100.0 km/h) to meters per second:
100.0 km/h * (1000 m/1 km) * (1 h/3600 s) = 27.78 m/s.

Since the final velocity (v) is 0 m/s and we know the distance (d) is 37.19 m, we can use the equation:
v^2 = u^2 + 2ad,
where u is the initial velocity, a is the acceleration, and d is the distance. Rearranging for a gives us:
a = (v^2 - u^2) / (2d) = (0 - 27.78^2) / (2 * 37.19) = -10.37 m/s².

The car's average acceleration during a stop is -10.37 m/s², which is a deceleration as indicated by the negative sign.