Decide whether the following problem can be solved using precalculus, or whether calculus is required. If the problem can be solved using precalculus, solve it. If the problem seems to require calculus, explain your reasoning and use a graphical or numerical approach to estimate the solution.
Find the distance traveled in 45 seconds by an object traveling at a velocity of v(t) = 20 + 3cos(t) feet per second. (Round your answer to the nearest foot.)
4 answers
distance=INTEGRAL velocity*dtime
In cos t I assume t is in degrees. If you mean radians, then do it that way.
It is a calculus problem because every little bit of distance is v(t) times a tiny bit of time, delta t. It v(t) were constant you could just multiply v times change in time to get the answer, but v is not constant.
A primitive way to do this is to split the 45 seconds up into three 15 second periods. Approximate that it goes v(0) for the first 15, then v(15) for the second 15 (15 to 30), then v(30 for the last 15 (30 to 45)
v(0)(15) = 23 * 15 = 345
v(15) * 15 = 22.89 * 15 = 343.5
v(30) * 15 = 22.6 * 15 = 339
so
d = 1027.5
Next divide it into 5 spaces of 9 seconds each and see how much the answer changes.
It is a calculus problem because every little bit of distance is v(t) times a tiny bit of time, delta t. It v(t) were constant you could just multiply v times change in time to get the answer, but v is not constant.
A primitive way to do this is to split the 45 seconds up into three 15 second periods. Approximate that it goes v(0) for the first 15, then v(15) for the second 15 (15 to 30), then v(30 for the last 15 (30 to 45)
v(0)(15) = 23 * 15 = 345
v(15) * 15 = 22.89 * 15 = 343.5
v(30) * 15 = 22.6 * 15 = 339
so
d = 1027.5
Next divide it into 5 spaces of 9 seconds each and see how much the answer changes.
I suspect that is not allowed Bob :) Too easy.
Find the distance traveled in 15 seconds by an object traveling
at a constant velocity of 20 feet per second.
at a constant velocity of 20 feet per second.
0 answers