# Variable acceleration

1. Aug 8, 2012

### monkfishkev

An object starts from position X = 0 and moves along a straight line with its position given by; X = 20t - 5t^3 where t is the time taken to reach position X.

Find;
i) The time at which the velocity is zero
ii) The time at which the acceleration is zero
iii) The average velocity over the first three seconds of motion

No answers were given with the questions, so would appreciate some answers to check against mine. I have 1.15s, 0s and -50m/s
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 8, 2012

### andrien

yes,yes, no(-25)

3. Aug 8, 2012

### monkfishkev

Can you please explain how you come to get -25. Thanks for your time :-)

4. Aug 8, 2012

### monkfishkev

I must be doing this wrong:

v = 20 - 15t^2
After 1s, v = 20 - 15(1) = 5m/s
After 2s, v = 20 - 15(4) = -40m/s
After 3s, v = 20 - 15(9) = -115m/s

Average speed = 5 - 40 - 115 =
= -150 ÷ 3 = -50m/s

5. Aug 8, 2012

### lewando

You are looking for average velocity. This is defined as displacement / time.

6. Aug 9, 2012

### andrien

avg. velocity=(∫vdt)/(∫dt),but here ∫vdt is x itself(given) so for three seconds
x=-75
t=3
<v>=-25

7. Aug 9, 2012

thank you