Practice 2C: | 1
| 2
| 3
| 4
| 5
| Go
up

**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 = ˝(v_{i} + v_{f})Dt

Δx =
˝(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 = ˝(v_{i} + v_{f})Dt
Δx = ˝(15 + 0)(2.5) Δx = **18.8 m**

**(Table
of contents)**

**3. A jet plane lands with a speed of 100 m/s and can
accelerate uniformly at a maximum rate of -5.0 m/s˛ 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/s˛, so 100 m/s ÷ 5 m/s˛ = 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 = ˝(v_{i} + v_{f})Dt
into Dt = Δx/ (˝(v_{i} + v_{f}))

Dt = 800/ (˝(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/ (˝(v_{i} + v_{f}))
so Dt = 99/ (˝(21.666 + 0)) **Dt
= 9.1 seconds**

**(Table
of contents)**

**5. A car enters the freeway with a speed of 6.4 m/s and
accelerates uniformly for 3.2 km in 3.5 min. How fast is the car moving after
this time? **

Make all of the units happy to start with:

3.2 km * 1000 m/km = 3200 m 3.5 min * 60 s/min = 210 s

Then, plug the values into the equation (you’re solving for final velocity)

3200 = ˝(6.4 + v_{f})(210) 3200/105
= 6.4 + v_{f }**v _{f
}= 24 m/s**