Question

A driver slows down her car from 32.7 km/hr at a constant rate of 0.63 m/s2 just by taking her foot of the accelerator (the gas pedal). Calculate the distance the car travels during the sixth second (between t=5 and t=6 seconds)that she is coasting.

Answer 1

Answer: She moves 5.616 meters in that second.

Explanation:

If we define t = 0s as the moment when she starts decelerating we can write the function of acceleration as:

a(t) = -(0.63 m/s^2)

where the negative sign is because she is slowing down.

The velocity equation can be found if we integrate over time:

v(t) = -(0.63m/s^2)*t + v0

Where v0 is the constant of integration, that represents the initial velocity, in this case is:

v0 = 32.7 km/h

Now, because the acceleration is in m/s^2, we should write this velocity in m/s.

in one km we have 1000 meters, and in one hour we have 3600 seconds, then we have that:

32.7 km/h = 32.7 *(1000/3600) m/s = 9.08 m/s

Then the velocity equation becomes:

v(t) = -(0.63m/s^2)*t  + 9.08 m/s

And for the position equation, we should integrate again to get:

p(t) = -(1/2)*(0.63m/s^2)*t^2 + (9.08m/s)*t + p0

Where p0 is the initial position.

For this problem, we want to find the distance that she moved between t = 5s and t = 6s, and that can be calculated as:

D = p(6s) - p(5s)

D = -(1/2)*(0.63m/s^2)*(6s)^2 + (9.08m/s)*6s + p0 +(1/2)*(0.63m/s^2)*(5s)^2 - (9.08m/s)*(5s) - p0

D = -(1/2)*(0.63m/s^2)*((6s)^2 - (5s)^2) + (9.08m/s)*(6s - 5s)

D = 5.615 m

She moves 5.616 meters in that second.