# Homework Help: Motion in a straight line and a cyclist

1. Sep 20, 2014

### Mary1910

Could someone please check my work? Thanks :)

A cyclist rode east for 1.0 x 10^2 m with a constant velocity of 25.0m/s. She then accelerated down a hill, and 5.0s later reached the bottom of the hill with a velocity of 50.0m/s.

a)How long did the cyclist ride before reaching the hill?

time=distance/velocity

t=1000m / 25m/s
=40s

b)What was her average acceleration down the hill?

acceleration=change in velocity/time

a=50m/s-25m/s / 5s
=25m/s / 5s
=10m/s^2

c)What was the distance she travelled down the hill?

distance=velocity x time

d=(50.0m/s)(5s)
=250m

d)If she continued to ride a constant velocity for another 5.0s, how far beyond the hill would she travel?

distance=velocity x time

d=(50.0m/s)(5s)
=250m

At the instant a traffic light turns green, a car starts from rest and accelerates uniformly at a rate of 4.0m/s (E). At the same instant, a truck travelling with a constant velocity of 90.0km/h (E) overtakes and passes the car.

a)How far beyond the starting point is the car after 10.0s?

delta d= (v1)(delta t) + 1/2 (a)(delta t^2)
=(0)(10)+1/2(4.0m/s)(100)
=200m

b)How far beyond the starting point is the truck after 10.0s?

distance=velocity x time
=(90km/h)(10.0s)
=(25m/s)(10.0s)
=250m

c)The car passes the truck at a distance of 312.5m beyond the starting point. How fast is the car travelling at this instant?

v2^2=v1^2 + 2(a)(delta d)
=0+2(4.0m/s)(312.5m)
=2500m/s^2
=50m/s

d)How long does the car take to pass the truck?

delta t=v2-v1/a
=50m/s-0m/s / 4.0m/s
=12.5s

e)Both drivers suddenly see a barrier 100.0m away and hit their brakes at exactly the same time. Assuming that both vehicles decelerate uniformly, and they take 3.0s to stop, will they stop in time?

Car
delta d=1/2(v1+v2)delta t
=1/2(50m/s)(3.0s)
=75m

Truck
delta d=1/2(v1+v2)delta t
=1/2(90km/h)(3.0s)
=1/2(25m/s)(3.0s)
=37.5s

Therefore both the car and the truck will be able to stop in time.

Thank you so much to anyone took the time to look over this and give feedback!

-Mary

2. Sep 21, 2014

### Simon Bridge

You have used a plug-and-chug style, which can trip you up when you pick the wrong equations - i.e. in part (c).
d=v/t only owrks for a constant velocity. While the cyclist was on the hill, she was accelerating.

The trick with these problems is to sketch the velocity-time graph.
The displacement is the area under the relevant section of the graph, the acceleration is the slope - that's usually easier than remembering the equations.

3. Sep 21, 2014

### Mary1910

Thank you

4. Sep 21, 2014

### Simon Bridge

No worries - how did you get on?