Post

AP Calculus AB: Velocity, Distance, and Direction

Created by @nathanedwards
 at October 31st 2023, 9:38:55 pm.
AP_Calculus_AB-Questions
#ap-calculus-ab
#position-velocity
#net-change

AP Calculus AB Exam Question

Suppose a cyclist is riding along a path and their position at any given time, t (in hours), is modeled by the function:

s(t)=3t2−4t+2s(t) = 3t^2 - 4t + 2

where s(t) is measured in meters.

a) Find the cyclist's velocity at any time t.

b) Calculate the total distance the cyclist travels from t = 0 to t = 2.

c) Determine the time(s) when the cyclist changes direction.

Solution

a) To find the cyclist's velocity at any time t, we need to take the derivative of the position function, s(t):

s(t)=3t2−4t+2s(t) = 3t^2 - 4t + 2

So, the velocity function, denoted as v(t), is:

v(t)=dsdt=ddt(3t2−4t+2)v(t) = \frac{ds}{dt} = \frac{d}{dt}(3t^2 - 4t + 2)

Differentiating each term of the function, we get:

v(t)=6t−4v(t) = 6t - 4

Therefore, the cyclist's velocity at any time t is given by the function v(t) = 6t - 4.

b) To calculate the total distance traveled by the cyclist from t = 0 to t = 2, we need to find the net change or accumulation of the absolute value of the velocity function over the interval [0, 2].

To do this, we integrate the absolute value of the velocity function with respect to t:

∫02∣v(t)∣dt=∫02∣6t−4∣dt\int_{0}^{2} |v(t)| dt = \int_{0}^{2} |6t - 4| dt

Since the velocity function can change sign, we need to split the integral at the point where it changes sign. In this case, the sign changes when 6t - 4 = 0, which occurs at t = 2/3. Thus, we can write the integral as:

∫02/3−(6t−4)dt+∫2/32(6t−4)dt\int_{0}^{2/3} -(6t - 4) dt + \int_{2/3}^{2} (6t - 4) dt

Simplifying, we have:

−∫02/3(6t−4)dt+∫2/32(6t−4)dt- \int_{0}^{2/3} (6t - 4) dt + \int_{2/3}^{2} (6t - 4) dt

Integrating each term separately, the calculation becomes:

=−[3t2−4t]02/3+[3t2−4t]2/32= - \left[3t^2 - 4t\right]_{0}^{2/3} + \left[3t^2 - 4t\right]_{2/3}^{2}

Plugging in the limits, we find:

=−[3(23)2−4(23)]−[3(2)2−4(2)]= - \left[3\left(\frac{2}{3}\right)^2 - 4\left(\frac{2}{3}\right)\right] - \left[3(2)^2 - 4(2)\right]
=−[2−83]−[12−8]= - \left[2 - \frac{8}{3}\right] - \left[12 - 8\right]
=−[63−83]−[4]= - \left[\frac{6}{3} - \frac{8}{3}\right] - \left[4\right]
=−[−23]−[4]= - \left[- \frac{2}{3}\right] - \left[4\right]
=23−4= \frac{2}{3} - 4
=−103= -\frac{10}{3}

Therefore, the total distance traveled by the cyclist from t = 0 to t = 2 is -10/3 meters.

c) To determine the time(s) when the cyclist changes direction, we need to find the time(s) when the velocity function, v(t), crosses the x-axis (i.e., when v(t) = 0).

Setting 6t - 4 = 0 and solving for t, we get:

6t−4=06t - 4 = 0
6t=46t = 4
t=46=23t = \frac{4}{6} = \frac{2}{3}

Thus, the cyclist changes direction at t = 2/3 hours.

Therefore, we have successfully answered all the questions related to net change and accumulation for this cyclist's motion.

Chat

Save To Bookmark