Position, Velocity, Acceleration, and Distance?

[Stanc]

Homework Statement

The position of a ball moving on a straight line has an equation of p(t) = 1/3t^3 - 2t^2 +3t , where p is its position and s is time in seconds

1. At what times is the ball at rest?

2. Using its velocity and acceleration, determine the balls DIRECTION and if it is SPEEDING UP or SLOWING DOWN at 2.5 seconds

3. Find the TOTAL DISTANCE the ball travels in the first 6 seconds.

The Attempt at a Solution

For 1, it is simple quadratics which I found the ball at rest to be at 0 seconds and also at 3 seconds

For 2, I believe that the first derivate equals velocity and the second derivative equals acceleration.

p'(t) = t^2 - 4t + 3
p"(t) = 2t - 4

For 2, would I be subbing in the time of 2.5s into my equation? If I do, I get a negative value which means that the velocity would be decreasing, correct? However, for the acceleration, I get a positive 1. Could someone please explain what that means? Would the ball also be moving in the straight linear direction still?

3. For this, I just subbed in 2.5 into first derivate which gave me 15m/s, I than multiplied by the 6 seconds and got 90 meters. Is this correct? Or Should i be subbing the 6 seconds into my original equation?

 

[rude man]

for 2, direction, determine dp/dt(t=2.5)
for speeding up, determine |dp/dt|(t=2.5)
for 3, distance = velocity integrated.

 

[Stanc]

Is what I did correct? Not quite sure what you are saying

 

[rude man]

Is what I did correct? Not quite sure what you are saying

1 is correct.
2 you didn't mention which equation you used. Also, I gave you wrong info on how to tell whether it's speeding up or slowing down: you got it right by substituting in for p''(2.5) I believe.
3. By "total distance covered" they may mean adding the negative parts of p. In other words, suppose you walk 10 ft. forwards, then 3 ft. back, then 8 ft. forwards again. "Total distance covered" might mean 21 ft or 15 ft. You interpreted it to mean the direct distance which is p(2.5) but as I said I suspect total distance covered includes the negative-going parts.

Anyway, you don't want to multiply the velocity at 2.5 s. by the time of 6s. That is a meaningless number. Substituting 6s in p is OK if they want the direct distance covered but as I said I think they mean to include the "backwards" parts of p also.

 

[Nickie]

I would like to clarify and provide some additional information:

1. The ball is at rest when its velocity is equal to zero. This occurs at t = 0 and t = 3 seconds, as you correctly stated.

2. To determine the direction of the ball and whether it is speeding up or slowing down at 2.5 seconds, you can use the first and second derivatives as you did. However, it is important to note that the first derivative gives the velocity at a specific time, while the second derivative gives the acceleration at that time. So, when you substitute t = 2.5 seconds into the first derivative, you get the velocity at that time, which is -2.75 m/s. This means that the ball is moving in the negative direction (since the original equation has a negative coefficient for the t^2 term) and is slowing down. When you substitute t = 2.5 seconds into the second derivative, you get the acceleration at that time, which is 0. This means that the ball is not accelerating at that time, it is at a constant velocity.

3. To find the total distance travelled in the first 6 seconds, you would need to integrate the absolute value of the first derivative (since the velocity can be both positive and negative) from t = 0 to t = 6. This would give you the displacement, which is the total distance travelled. Alternatively, you can use the distance formula d = |p(6) - p(0)|, where p(6) is the position at t = 6 seconds and p(0) is the position at t = 0 seconds. Both methods should give you the same answer. Substituting t = 6 into the original equation gives you a position of 9 meters, so the total distance travelled in the first 6 seconds is 9 meters.