## rvc one year ago The motion of a particle along a straight line is governed by the relation a= t^3-2t+7 where a is in m/s^2 and t is time in seconds. At time t=1 the velocity of the particle is 3.58m/s and displacement is 9.39m. Calculate the displacement,velocity and acc at t=2s. I know the formula but the calculation part is getting wrong somewhere.

1. Michele_Laino

you have to integrate that function in order to find the velocity v(t): $\Large v\left( t \right) = \int {\left( {{t^3} - 2t + 7} \right)dt}$

2. rvc

v = $\rm v=\frac{ t^4 }{ 4 } -2\frac{ t^3 }{ 3 }+7t+C$

3. rvc

brb

4. Michele_Laino

ok! now you have to substitute your condition, namely t=1 and v=3.58, so you will find the value of the constant C

5. Michele_Laino

I got this: $\Large v\left( t \right) = \int {\left( {{t^3} - 2t + 7} \right)dt} = \frac{{{t^4}}}{4} - {t^2} + 7t + C$

6. rvc

yes

7. Michele_Laino

after a substitution, you will get this: $\Large 3.58 = \frac{1}{4} - 1 + 7 + C$

8. rvc

$\sf 3.58=\frac{ 1 }{ 4 }-\frac{ 2 }{ 3 } +7+C$

9. Michele_Laino

why 2/3?

10. rvc

OHHHHHHHHHHHHHHHHHHhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh i got it correctly now because i copied the question wrong

11. Michele_Laino

is a(t) like this: $\Large a\left( t \right) = {t^3} - 2t + 7$