Thread: Help please, acceleration/velocity question.

The acceleration of a particle moving along a straight line is given by a=10e^2t.
a.) Write an expression for the velocity v, in terms of time t, if v=5 when t=0
b.) During the time when the velocity increases from 5 to 15 how far does the particle travel?
c. Write an expression for the position s, in terms of time t, of the particle if s=0 when t=0

I don't even know how to start the problem if someone can please help me, and explain to me how you do the problem I would greatly appreciate it.

Thanks.

Originally Posted by Jasmina8

The acceleration of a particle moving along a straight line is given by a=10e^2t.
a.) Write an expression for the velocity v, in terms of time t, if v=5 when t=0
b.) During the time when the velocity increases from 5 to 15 how far does the particle travel?
c. Write an expression for the position s, in terms of time t, of the particle if s=0 when t=0

I don't even know how to start the problem if someone can please help me, and explain to me how you do the problem I would greatly appreciate it.

Thanks.

here's how to start ...

velocity is the antiderivative of acceleration

position is the antiderivative of velocity

use your given initial conditions to determine the constant of integration in each case.

So because a=10e^2t the antiderivative of that would be:

a.)
v= ((10e^2t)/2)+C
5= ((10e^0)/2)+C
5=5+C
C=0
so v= 5e^2t?? I'm kind of confused about this so I'm not sure of my answer.

and I have no clue how to do b and c; for B I was thinking of solving the v=5e^2t for t when v=15 so
15=5e^2t/2
15=5e^0/2
15=2.5 which doesn't make any sense...

Originally Posted by Jasmina8

The acceleration of a particle moving along a straight line is given by a=10e^2t.
a.) Write an expression for the velocity v, in terms of time t, if v=5 when t=0
b.) During the time when the velocity increases from 5 to 15 how far does the particle travel?

a)



b) displacement