Analyzing motion problems: position | AP Calculus AB | Khan Academy

Divya received the following problem: A particle moves in a straight line with velocity ( v(t) ) is equal to the square root of ( 3t - 1 ) meters per second, where ( t ) is time in seconds. At ( t = 2 ), the particle's distance from the starting point was eight meters in the positive direction. What is the particle's position at ( t = 7 ) seconds? Which expression should Divya use to solve the problem?

So pause this video and have a go at it all right now.

Let's do this together! So we want to know the particle's position at ( t = 7 ). They tell us what our position is at ( t = 2 ). Thus, the position at ( t = 7 ) would be your position at ( t = 2 ) plus your change in position from ( t = 2 ) to ( t = 7 ).

There's another word for this; you could also call this your displacement from ( t = 2 ) to ( t = 7 ). We know how to think about displacement: velocity is your rate of change of displacement. If you want to figure out your displacement between two times, you would integrate the velocity function.

So this is going to be the integral from ( t = 2 ) to ( t = 7 ) of our velocity function ( v(t) , dt ). This would be our displacement from time ( 2 ) to time ( 7 ). If they asked what our change in position from time ( 2 ) to time ( 7 ) is, it would be just this expression.

But they want us, or they want Divya, to figure out what the particle's position is at ( t = 7 ) seconds. So what you would want to do is take your position at ( t = 2 ). We know what our position at ( t = 2 ) is; it was 8 meters in the positive direction, so we could just call that positive 8 meters.

Therefore, it’s going to be ( 8 ) plus your change in position, which is going to be your displacement. We can see this choice right over there, and that’s what we would pick.

The first option, ( v(7) ), just gives us our velocity at time ( 7 ) or, exactly at ( 7 ) seconds, or in other words, our rate of change of displacement at ( 7 ) seconds. So that’s not what we want.

The second option shows your position at ( t = 2 ), but then you have your change in position from ( t = 0 ) to ( t = 7 ), so this doesn’t seem right.

Lastly, this is your position at time ( 2 ) plus ( v' ), the derivative of velocity, which is the acceleration, plus your acceleration at time ( 7 ). So that's definitely not going to give you the particle’s position. We like that second choice.