Hi, Can you help explain how to obtain the maximum UPWARDS velocity of this grap

Question

Hi,

Can you help explain how to obtain the maximum UPWARDS velocity of this graph?

http://imgur.com/sTpZYUJ

We haven't covered derivatives in class yet, we've only averaged out the velocities from the given data to estimate an instaneous rate of change.

It looks like between seconds 3 - 3.5 look the steepest.  I've averaged out the velocity between 3 - 3.5 secs and it is 123 ft/ sec.  But the professor that isn't correct. (He says it is much higher)

Any advice would be helpful.

Thanks!

Can you help explain how to obtain the maximum UPWARDS velocity of this graph? We haven't covered derivatives in class yet, we've only averaged out the velocities from the given data to estimate an instaneous rate of change. It looks like between seconds 3 - 3.5 look the steepest. I've averaged out the velocity between 3 - 3.5 secs and it is 123 ft/ sec. But the professor that isn't correct. (He says it is much higher)

Explanation / Answer

Velocity is defined as the rate of change of distance i.e. v = dx/dt. So, if x vs. t data is given you need to find the slope of the curve which is equal to velocity. Maximum velocity occurs where slope is maximum

In this case you have t vs. h data. So, compute velocity as: v = (h2 - h1)/(t2 - t1) for every successive times and see when it is max.

So, you get something like this: Maximum occurs at t = 3s when velocity is 123 ft/s

0 0 12.8 0.5 6.4 34.2 1 23.5 55.8 1.5 51.4 77.2 2 90 98.8 2.5 139.4 120.2 3 199.5 123 3.5 261 107 4 314.5 91 4.5 360 75 5 397.5 59 5.5 427 43 6 448.5 27 6.5 462 11 7 467.5 -935

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.