SIMPLE Calculus Question

[math q]

Why on earth is this wrong?

v(t) = 6t^2-48t+90
a(t) = 12t-48

Solving a(t) for 0 should give what I put, right? Any suggestions?

Thanks in advance.

 

[Jason76]

View attachment 4570

Why on earth is this wrong?

v(t) = 6t^2-48t+90
a(t) = 12t-48

Solving a(t) for 0 should give what I put, right? Any suggestions?

Thanks in advance.

Perhaps the type of bracket "[ or )" is wrong. Just a guess. Anyone on the forum agree? However, your choice of brackets is similar to the velocity problem, which is correct by your online homework.

 

[HallsofIvy]

No, you have calculated the derivatives correctly and where the velocity and acceleration are positive and negative but you have misunderstood "speeding up" and "slowing down" The object will be "speeding up" as long as the velocity and acceleration are in the same direction- that is, where they both have the same sign regardless whether that sign is negative or positive. And it will be "slowing down" when they are opposite.

 

[Ishuda]

To say it a bit differently than HallsofIvy did - what is the speed when the particle turns around and starts going backward at 3 and again when it turns around and starts going forward at 5. Which direction was it going each time?

 

[Jason76]

It would be speeding up from \(\displaystyle (4,\infty)\) cause same sign for velocity and acceleration there. On 4 the acceleration and velocity are same sign (negative) Past 4, acceleration and velocity are same sign (positive)

Slowing down on \(\displaystyle [0,3)\) cause different signs for velocity and acceleration.

 

[math q]

Awesome, I figured it out! Thanks to all that posted! . As you guys mentioned, it was because I falsely interpreted what speeding up and speeding down meant.