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lisov135 [29]
3 years ago
5

A long distance runner sees the finish line and accelerates at a rate in 1.2 m/s2 for

Physics
1 answer:
Nuetrik [128]3 years ago
6 0

Answer: he did travel 15 meters.

Explanation:

We have the data:

Acceleration = a = 1.2 m/s^2

Time lapes = 3 seconds

Initial speed = 3.2 m/s.

Then we start writing the acceleration:

a(t) = 1.2 m/s^2

now for the velocity, we integrate over time:

v(t) = (1.2 m/s^2)*t + v0

with v0 = 3.2 m/s

v(t) = (1.2 m/s^2)*t + 3.2 m/s

For the position, we integrate again.

p(t) = (1/2)*(1.2 m/s^2)*t^2 + 3.2m/s*t + p0

Because we want to know the displacementin those 3 seconds ( p(3s) - p(0s)) we can use p0 = 0m

Then the displacement at t = 3s will be equal to p(3s).

p(3s) = (1/2)*(1.2 m/s^2)*(3s)^2 + 3.2m/s*3s = 15m

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CHECK THE ATTACHMENT

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