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3 years ago

A car can move from rest to 35 m/s in 5.0 seconds. It's acceleration is

Physics

1 answer:

lord [1]3 years ago

6 0

Answer:

A) 7.0 m/s²

Explanation:

Acceleration is found by using the formula:

a = Δv/Δt

This means the change in time divided by the time interval will give us the car's acceleration.

Since the car starts from rest, its initial velocity is 0 m/s. Its final velocity is 35 m/s. The time interval for the car to go from 0 to 35 m/s is 5.0 seconds.

a = (v - v₀)/t

Substitute the final and initial velocity into the formula.

a = (35 - 0)/5
a = 35/5
a = 7

The acceleration of the car is A) 7.0 m/s².

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The relative velocity of the athlete relative to the ground is 5.2 m/s

The given parameters;

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let the velocity of the ground = Vg = 0

The relative velocity concept helps us to determine the velocity of a moving object relative to a stationary observer.

The athlete is the moving object in this question while the ground is stationary.

The relative velocity of the athlete relative to the ground is calculated as follows;

$V/V_g = V - V_g = 5.2 - 0 = 5.2 \ m/s$

Thus, the relative velocity of the athlete relative to the ground is 5.2 m/s

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