You are driving at 60 mph you slow down to a stop which takes 200 meters..

Question

.. Then you speed up to 30 mph which took 70 seconds

1. how much time did this trip take?
2. how far did you drive total?

*convert units to m/s

Answer 1

The whole trip consists of two portions -- before the stop and after the stop.

For both portions we know the initial and final velocities, hence the change in velocity Δv.

For the first portion we know the distance Δs, but not the time Δt.

For the second portion we know the time Δt, but not the distance Δs.

To solve this problem, we need to assume that during both portions the car

undergoes constant acceleration (deceleration), a.

We analyze both portions separately, solving for the unknown quantity.

For both portions we use the identities

a = Δv/Δt (1)

Δs = aΔt²/2 (2)

Substituting a from the first equation into the second:

Δs = ΔvΔt/2 (3)

We can use (3) to find the unknown Δs in the second portion.

To find the unknown Δt in the first portion we rewrite (3) into

Δt = 2Δs/Δv (4)

FIRST PORTION:

Δv = 60mph = 60×1600m/3600s = 80/3 m/s

Δs = 200m

Using (4)

Δt = 2×200m/(80/3 m/s) = 15s

SECOND PORTION:

Δv = 30mph = 30×1600m/3600s = 40/3 m/s

Δt = 70s

Using (3)

Δs = (40/3 m/s × 70 s)/2 = 467 m

FINAL RESULT:

The trip took 15s + 70s = 85s

You drove 200m + 467m = 667m