Previous Calculus AB AP test question

  • Thread starter Thread starter CRW1688
  • Start date Start date
  • Tags Tags
    Ap Calculus Test
CRW1688
Messages
1
Reaction score
0

Homework Statement



t 0 10 20 30 40 50 60 70 80
v(t) 5 14 22 29 35 40 44 47 49

This problem is from a previous Calculus AP test. I have completed parts (a) and (c), but (b) is giving me trouble, so I will leave parts (a) and (c) off.

Rocket A has positive velocity v(t) after being launched upward from an initial height of 0 feet at time t=0 seconds. The velocity of the rocket is recorded for selected values of t over the time interval 0 <= t <= 80 seconds, as shown in the table above.

Using correct units, explain the meaning of the integral of v(t) dt from 10 to 70. Use a midpoint Riemann sum with three subintervals of equal length to approximate the integral of v(t) dt form 10 to 70.



2. The attempt at a solution

The teacher took away our books, and we're not supposed to use calculators. My Riemann sum memory is a little weak, but I worked the second part of the question to the best of my ability, using three Riemann sums, one from 10 to 30 with midpoint 20, one from 30 to 50 with midpoint 40, and one from 50 to 70 with midpoint 60.

I found the area of the rectangles created by the intervals by calculating the width * the height at the midpoint, thus leaving me with:

M3=20(22) + 20(35) + 20(44) = 2020

I may still be way off base, as I can't look up how to do Riemann sums in the book (I could look online, but I want to see if I'm on the right track first).


The first part of the question, the "explain the meaning" part, is mostly what I don't get. Any help?
 
Physics news on Phys.org
velocity is the derivative of position.. i.e.
v(t)=\frac{dx}{dt}

so, when you integrate the velocity you get
\int_a^b v(t)dt=\int_a^b \frac{dx}{dt}dt=\int_a^bdx=x(t)|_a^b

In other words, when you take the integral of the velocity over a period of time you are essentially solving for the total distance traveled over that period of time.
 
what year was this question asked?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Back
Top