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Suppose a car accelerates with an average acceleration of 20.0. m/s² starting from rest for 5 sec. How far does it travel at this time?

User DYS
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Final answer:

The distance traveled by a car accelerating from rest at an acceleration of 20.0 m/s² for 5 seconds is 250 meters, calculated using the kinematic equation d = ½at².

Step-by-step explanation:

To calculate the distance traveled by a car that accelerates from rest with a given average acceleration, we can use the kinematic equation for motion that relates distance (d), initial velocity (v0), acceleration (a), and time (t):

d = v0t + ½at²

Since the car starts from rest, its initial velocity is 0 m/s. Given an average acceleration of 20.0 m/s² and a time of 5 sec, we can substitute into the equation:

d = 0 × 5 + ½ × 20.0 × (5)²

d = ½ × 20.0 × 25

d = 10.0 × 25

d = 250 m

Therefore, the car travels a distance of 250 meters during the 5 seconds of acceleration.

User Rich Steinmetz
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