Position, Velocity, Acceleration, and Distance?

AI Thread Summary
The discussion focuses on a homework problem involving the motion of a ball described by the position equation p(t) = 1/3t^3 - 2t^2 + 3t. The ball is at rest at 0 and 3 seconds, determined by solving the quadratic equation derived from the first derivative. At 2.5 seconds, the velocity is negative, indicating the ball is slowing down, while the positive acceleration suggests it is still moving in a straight line. For total distance traveled in the first 6 seconds, it is clarified that the calculation should consider both forward and backward movement, rather than simply multiplying velocity by time. The correct approach involves integrating the velocity to account for all movement directions.
Stanc
Messages
60
Reaction score
0

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?
 
Last edited:
Physics news on Phys.org
for 2, direction, determine dp/dt(t=2.5)
for speeding up, determine |dp/dt|(t=2.5)
for 3, distance = velocity integrated.
 
Is what I did correct? Not quite sure what you are saying
 
Stanc said:
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.
 
Last edited:
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top