A motorcyclist who is moving along an x-axis has an acceleration given by a=(6.60-2.29t) for 0 >= t <= 6.0 s, where a is in m/s2 and t is in seconds. At t = 0, the velocity and position of the cyclist are 3.20 m/s and 7.30 m. For the time interval from 0 >= t <= 6.0 s, what is the difference between the maximum speed and the average speed of the motorcyclist?
I tried finding the position and velocity functions and plugging in 6secs for both functions to find total distance and max speed, and then finding the average velocity, but when I subtracted the maximum speed and the average velocity I keep getting the wrong answers. The answer was supposed to be 3.45m/s, and I'm always off by a few numbers.
a(t) = 6.60-2.29t
v(t) = 6.60t - 1.145t^2 + c
v(0) = 3.20, so c = 3.20, and
v(t) = 3.20 + 6.60t - 1.145t^2
s(t) = 3.20t + 3.30t^2 - 0.3817t^3 + c
s(0) = 7.30, so c = 7.30
s(t) = 7.30 + 3.20t + 3.30t^2 - 0.3817t^3
max speed occurs when a=0, or at t=2.882
v(2.882) = 12.7109
s(6.0) = 62.8528
avg speed is s(6)/6 = 10.4755
vmax - vavg = 2.235 m/s
If your answer agrees with this (should have shown your work), then we can be pretty sure the supposed answer is bogus or there's a typo in the problem.
If we disagree, double-check my math and yours.
To find the difference between the maximum speed and the average speed of the motorcyclist, we need to follow a few steps. Let's break it down:
Step 1: Find the velocity function (v) by integrating the acceleration function (a) with respect to time (t).
Given: a = 6.60 - 2.29t
To find v, integrate a with respect to t:
v = ∫(6.60 - 2.29t) dt
When you integrate, the constant of integration arises. Since we are given the initial velocity at t = 0 as 3.20 m/s, we can solve for the constant of integration (C):
3.20 = ∫(6.60 - 2.29(0)) dt
3.20 = 6.60t - 1.145t^2 + C
Now, we have the velocity function:
v = 6.60t - 1.145t^2 + C
Step 2: Find the position function (x) by integrating the velocity function (v) with respect to time (t).
Given: v = 6.60t - 1.145t^2 + C
To find x, integrate v with respect to t:
x = ∫(6.60t - 1.145t^2 + C) dt
When you integrate, another constant of integration arises. Since we are given the initial position at t = 0 as 7.30 m, we can solve for the new constant of integration (D):
7.30 = ∫(6.60t - 1.145t^2 + C) dt
7.30 = 3.30t^2 - 0.573t^3 + Ct + D
Now, we have the position function:
x = 3.30t^2 - 0.573t^3 + Ct + D
Step 3: Calculate the maximum speed.
To find the maximum speed, we need to find the maximum value of the velocity function. We can do this by taking the derivative of the velocity function (v) with respect to time (t) and setting it to zero:
dv/dt = 6.60 - 2.29t
Set dv/dt = 0:
6.60 - 2.29t = 0
t = 6.60 / 2.29
t ≈ 2.88 s
Now, substitute this value of t into the velocity function (v) to find the maximum speed:
v = 6.60(2.88) - 1.145(2.88)^2 + C
Step 4: Calculate the average speed.
Average speed is defined as the total distance traveled divided by the total time taken. To find the average speed, we need to find the total distance traveled and the total time taken.
Total distance traveled:
Since we already have the position function (x), substitute t = 6.0 s into the position function:
x = 3.30(6.0)^2 - 0.573(6.0)^3 + C(6.0) + D
Total time taken: 6.0 s
Average speed = Total distance / Total time
Step 5: Calculate the difference between the maximum speed and the average speed.
Difference = Maximum speed - Average speed
Now, put all the obtained values together to calculate the final answer.