I won't do that again.Problem: Integrating Kinematic Equations to Find Position

[jjiimmyy101]

Problem: A car starts from rest and moves along a straight line with an acceleration of a=(3s^-1/3)m/sec^2, where s is in metres. Determine the car's acceleration when t=4sec. ANS: 1.06 m/sec^2

Alright...I know nothing about integrals...really, nothing. I was never taught anything about integrals even though I've taken calculus courses before.

Here's what I think I should do.

Take the equation a = d^2s/dt^2 and INTEGRATE it to find the position (s). But how do you do this.

 

[pnaj]

Forget numbers for a minute.

If ao is acceleration, vo is initial velocity, xo is initial position, then,

Acceleration at time t is: a(t) = ao
Velocity at time t is: v(t) = vo + ao * t
Position at time t is: x(t) = xo + vo * t + (ao * t^2)/2


They haven't taught you integration yet. Maybe you've been given these formulae, though, or something similar.

 

[jjiimmyy101]

what's the initial position (xo)?

 

[pnaj]

You don't need to know to answer the question. It's the distance, d, that's needed ...
d = x(4) - xo
= ...

 

[jjiimmyy101]

d = 24s^-1/3

Sorry, I'm still not understanding.
I really appreciate your help, though.

 

[pnaj]

How far you go doesn't depend on where you start from, so take xo to be as simple as possible ... xo = 0.

 

[himanshu121]

Originally posted by pnaj
Forget numbers for a minute.

If ao is acceleration, vo is initial velocity, xo is initial position, then,

Acceleration at time t is: a(t) = ao
Velocity at time t is: v(t) = vo + ao * t
Position at time t is: x(t) = xo + vo * t + (ao * t^2)/2


They haven't taught you integration yet. Maybe you've been given these formulae, though, or something similar.

The equations are valid only when a is constant

 

[pnaj]

Thanks for that, himanshu121, I didn't notice the 's' sitting in the expression for a.

I just assumed that it was just a simple problem that didn't need integration and got side-tracked into seeing if jimmy had seen these equations before.