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Displacement with uniform acceleration

- by the terrific twins Heather and Lisa, 2005

1. A car accelerates uniformly from rest to a speed of 23.7 km/h in 6.5 s. Find the distance the car travels during this time.

First, change 6.5 seconds into hours                  6.5s * h/3600s =  .0018055 hours

Then, plug it in to the equation Δx = ½(vi + vf)Dt

Δx =  ½(0 + 23.7)(.0018055) = .021 km

2. When Maggie applies the brakes of her car, the car slows uniformly from 15.0 m/s to 0m/s in 2.50 s. How many meters before a stop sign must she apply her brakes in order to stop at the sign?

Δx = ½(vi + vf)Dt          Δx = ½(15 + 0)(2.5)    Δx = 18.8 m

3. A jet plane lands with a speed of 100 m/s and can accelerate uniformly at a maximum rate of -5.0 m/s² as it comes to rest. Can this plane land at an airport where the runway is .80 km long?

You can start by finding out how much time the plane would need to land. It’s going 100 m/s and decelerates  at 5.0 m/s², so             100 m/s ÷ 5 m/s² = 20 seconds

Now, find how long the plane has to stop on a .8 km long runway

.8 km * 1000 m/km = 800 m

You’re solving for Dt so change Δx = ½(vi + vf)Dt  into  Dt = Δx/ (½(vi + vf))

Dt = 800/ (½(100 + 0))            Dt = 16 seconds

The plane needs 20 seconds to decelerate completely but it will only get 16 seconds on the runway so the answer is no, it can’t land at the airport.

4. A driver in a car traveling at a speed of 78 km/h sees a cat 101 m away on the road. How long will it take for the car to accelerate uniformly to a stop in exactly 99 m?

First, change km/h into m/s   78 km/h * 1000 m/km = 78000 m/h * h/3600 s = 21.666 m/s

Dt = Δx/ (½(vi + vf))  so  Dt = 99/ (½(21.666 + 0))      Dt = 9.1 seconds