1. Jun 12, 2014

### risingabove

1. The problem statement, all variables and given/known data

A bus is travelling steadily at 30 m/s along a straight road passes a stationary car which, 5s later, begins to move with a uniform acceleration of 2 m/s^2 in the same direction as the bus.

(i) how long does it take the car to acquire the same speed as the bus?

(ii) how far has the car travelled when it is level with the bus?

2. Relevant equations

v=u+at
v^2 = u^2 + 2as
s= ut + 1/2 (a)( t^2)
s = 1/2 (u+v)t

where by

s= distance
u = initial velocity
v=final velocity
a= acceleration
t= time

3. The attempt at a solution

Attempt at part (i)

I plug in the values into v=u+at

using v = 30m/s
u = 0m/s
a = 2m/s^2

therefore 30 = 0 + 2 (t)
t = 15s

Attempt at part (ii)

I plug in the values into s= ut + 1/2 (a)( t^2)

using u=30m/s
a= 2m/s^2
t = 20 s i got this value by adding the 15s i got from the previous working to the 5s from the question

therefor s = 30(20) + 1/2 (2) (20)^2
S = 1000m

However the answer that was given was 1181m

i cant figure out where did i go wrong, or what step i am missing....Please help me...greatly appreciate it
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 12, 2014

### Nathanael

This is where your mistake is.

15s is the time it takes to get up to speed with the bus. This has nothing to do with getting up to the position of the bus.

You're going to want to set up an equation that involves the position of the bus after t seconds and the position of the car after t seconds then set them equal and solve for t

3. Jun 12, 2014

### BvU

Sorry to barge in like this, but there is something terribly wrong with
u is for the bus
a is for the car
The equation works for linear motion with constant acceleration. during the 5 s a = 0m/s2, then it is 2 m/s2. So the sum can never be used in the equation.
So s calculated in this way is meaningless2.

This well-meant reply in partial answer to "where did I go wrong".

4. Jun 12, 2014

### risingabove

why am i solving for t? and after t seconds... am i using 5 seconds? what t seconds?

5. Jun 12, 2014

### risingabove

No reason to apologize, didnt seem as if you barge in, however with your constructive advice i would of liked some form of help in regards to correcting where i did go wrong as you have pointed out.

6. Jun 12, 2014

### Nathanael

There are two functions involved, the position of the car (as a function of time) and the position of the bus (as a function of time)

If you graphed these functions together on the same x-y coordinate system (x is time, y is position) then the intersection of the two graphs would be the solution of this problem (because you want to find when the car and bus are at the same position at the same time)

The reason you need to solve for the time is that the bus and car have to be next to eachother at the same time

You don't need to graph it, though. You can create an equation that describes the position of the bus at any time t (Here t is a variable, it does not have a specific value. Call it x if you want.)
And then you can set it equal to the equation that desribes the position of the car at any time t
(You can set it equal because the position of the car is equal to the position of the bus)

Then you can solve for how long it takes for the car to catch up to the bus (that would be t) and then you can plug t back into either equation to find the distance

7. Jun 12, 2014

### BvU

Fair enough.

Reason you "need" t is because s and t are related through the equations for the distances travelled by car and bus. Both are unknown as yet, but you have one eqn for the car and one for the bus.

Richard (oops, Nathaniel) means: eliminate t. Solving for t is almost equivalent, since one of the equations is so simple.

 oops, typing to slowly. Must be bedtime. Over to you, Nat !

8. Jun 12, 2014

### Nathanael

Hahaha this gave me a laugh :)

9. Jun 13, 2014

### risingabove

Thank much for the help:) i followed the directions given and got it worked out...:)