# Homework Help: Kinematics help

1. Aug 26, 2011

### Goopy17

1. The problem statement, all variables and given/known data
The space shuttle undergoes an acceleration of 53.9 m/s^2. How fast is it traveling at the end of 55.2 s?

2. Relevant equations

3. The attempt at a solution
I have no clue. I am doing a summer assignment and it is very poorly explained by the teacher.

2. Aug 26, 2011

### PeterO

Are you familiar with equations summarising motion with constant acceleration?
An example of the set is

v = u + at
s = ut + (1/2).a.t^2
s = vt - (1/2).a.t^2
v^2 = u^2 +2as
s = [(u+v)/2]*t

some people/texts use different symbols

3. Aug 26, 2011

### Goopy17

What is u? v is velocity, a is acceleration, and t is time, but what is u?

4. Aug 26, 2011

### PeterO

u is the initial velocity, v is the final velocity [like in the alphabet u comes before v]

s is displacement

with those symbols you don't need subscripts on the letters

5. Aug 26, 2011

### Goopy17

Could you help me work it out?

6. Aug 26, 2011

### Goopy17

What is the value for initial velocity?
V= 53.9 m/s^2
A= 53.9m/s^2
T= 55.2s
S= ?
I got distance as 2975.28 but I don't know if that is correct

7. Aug 26, 2011

### Goopy17

Or are we looking for velocity?

8. Aug 26, 2011

### PeterO

Firstly you identify the value of the three properties you know - like u = 0

Then you identify which variable you are trying to find. That part is easy. You find the question mark in the question, then move back through the text until you find the previous punctuation mark. Then you read the words between that punctuation mark and the question mark.

You then choose the equation that connects those 4 variables, substitute and solve.

9. Aug 26, 2011

### Goopy17

So we are looking for speed, right?

10. Aug 26, 2011

### Goopy17

I got 2975.28 m/s, but I am not sure if that is correct.

11. Aug 27, 2011

### PeterO

Which formula did you use?

12. Aug 27, 2011

### PeterO

Given that we have u = initial speed, & v = final speed you should be more specific, but yes we are looking for speed.

13. Aug 27, 2011

### PeterO

Why were you not sure it was correct?