Connecting Position, Velocity, and Acceleration
Help Questions
AP Calculus BC › Connecting Position, Velocity, and Acceleration
A motorcycle’s acceleration is $a(t)=2\cos t$ with $v(0)=-1$; what is $v\left(\frac{\pi}{2}\right)$?
$v\left(\frac{\pi}{2}\right)=\int_0^{\pi/2}2\cos t,dt$
$v\left(\frac{\pi}{2}\right)=-1+\int_0^{\pi/2}\cos t,dt$
$v\left(\frac{\pi}{2}\right)=-1+2\cos\left(\frac{\pi}{2}\right)$
$v\left(\frac{\pi}{2}\right)=-1+\int_0^{\pi/2}(-2\sin t),dt$
$v\left(\frac{\pi}{2}\right)=-1+\int_0^{\pi/2}2\cos t,dt$
Explanation
This problem involves using integration in kinematics to connect acceleration to velocity. To find the velocity at t=π/2, integrate the acceleration function a(t)=2 cos t from 0 to π/2, which gives the change in velocity over that interval. Then, add the initial velocity v(0)=-1 to obtain the final velocity. This results in v(π/2)=-1 + ∫ from 0 to π/2 of 2 cos t dt, where the integral represents the net change due to acceleration. A tempting distractor is choice C, which adds the acceleration at t=π/2 to the initial velocity, mistaking instantaneous acceleration for the total change. In general, when solving initial value problems in kinematics, determine the constant of integration by applying the initial condition after finding the antiderivative.
A cyclist’s acceleration is $a(t)=4\cos(2t)$ and $v(0)=1$. What is $v\left(\dfrac{\pi}{4}\right)$?
$-1$
$-3$
$3$
$1$
$5$
Explanation
This problem involves the kinematics skill of integrating acceleration to find velocity. To find the velocity function v(t), we integrate a(t) = 4 cos(2t) to obtain v(t) = 2 sin(2t) + C. Using the initial condition v(0) = 1, we substitute t = 0 to get C = 1. Thus, v(π/4) = 2 sin(π/2) + 1 = 2*1 + 1 = 3. A tempting distractor is 1, from forgetting to integrate and just using the initial value. Always integrate the given function and use the initial condition to find the constant of integration, then evaluate at the desired point.
A particle has velocity $v(t)=3t^2-4t+1$ for $0\le t\le2$ and position $s(0)=5$; find $s(2)$.
$\int_0^2 (3t^2-4t+1),dt$
$11+5$
$5+v(2)$
$5+\int_0^2 (3t^2-4t+1),dt$
$11$
Explanation
This problem requires using kinematics integration to find position from velocity. Since velocity v(t) = 3t² - 4t + 1 is the derivative of position, we find position by integrating: s(t) = s(0) + ∫₀ᵗ v(u)du. With s(0) = 5, we get s(2) = 5 + ∫₀² (3t² - 4t + 1)dt. Choice C incorrectly omits the initial position s(0) = 5, giving only the displacement from t = 0 to t = 2. When solving position-velocity problems, always remember to add the initial condition to the integral of velocity.
A particle has velocity $v(t)=\sin t+t$ for $0\le t\le \pi$ and $s(0)=-1$; find $s(\pi)$.
$s(\pi)=-1+(\sin \pi+\pi)$
$s(\pi)=-1+\int_0^{\pi}(\sin t),dt$
$s(\pi)=-1+\int_0^{\pi}(\sin t+t),dt$
$s(\pi)=\int_0^{\pi}(\sin t+t),dt$
$s(\pi)=-1+\int_0^{\pi}(\cos t+1),dt$
Explanation
This problem involves using integration in kinematics to connect velocity to position. To find the position at t=π, integrate the velocity function v(t)=sin t + t from 0 to π, which gives the change in position over that interval. Then, add the initial position s(0)=-1 to account for the starting point. This yields s(π)=-1 + ∫ from 0 to π of (sin t + t) dt, as the integral computes the net displacement. A tempting distractor is choice C, which evaluates the velocity at t=π and adds it to the initial position, mistaking velocity for displacement. In general, when solving initial value problems in kinematics, determine the constant of integration by applying the initial condition after finding the antiderivative.
A drone’s acceleration is $a(t)=\frac{6}{(t+1)^2}$ with $v(2)=4$; what is $v(5)$?
$v(5)=4+\int_0^5\frac{6}{(t+1)^2},dt$
$v(5)=4+\int_2^5\frac{6}{t+1},dt$
$v(5)=\int_2^5\frac{6}{(t+1)^2},dt$
$v(5)=4+\frac{6}{(5+1)^2}$
$v(5)=4+\int_2^5\frac{6}{(t+1)^2},dt$
Explanation
This problem involves using integration in kinematics to connect acceleration to velocity. To find the velocity at t=5, integrate the acceleration function a(t)=6/(t+1)² from 2 to 5, which gives the change in velocity over that interval. Then, add the initial velocity v(2)=4 to obtain the final velocity. This results in v(5)=4 + ∫ from 2 to 5 of 6/(t+1)² dt, ensuring the limits align with the initial condition. A tempting distractor is choice D, which adds the acceleration at t=5 to the initial velocity, confusing instantaneous values with integrated change. In general, when solving initial value problems in kinematics, determine the constant of integration by applying the initial condition after finding the antiderivative.
A car’s acceleration is $a(t)=4t-6$ (m/s$^2$) with $v(0)=5$; what is $v(2)$?
$v(2)=5+\int_0^2(4t-6)',dt$
$v(2)=5+\int_0^2(4-6t),dt$
$v(2)=5+\int_0^2(4t-6),dt$
$v(2)=\int_0^2(4t-6),dt$
$v(2)=5+(4\cdot2-6)$
Explanation
This problem involves using integration in kinematics to connect acceleration to velocity. To find the velocity at t=2, integrate the acceleration function a(t)=4t-6 from 0 to 2, which gives the change in velocity over that interval. Then, add the initial velocity v(0)=5 to obtain the final velocity. This results in v(2)=5 + ∫ from 0 to 2 of (4t-6) dt, where the integral represents the net change due to acceleration. A tempting distractor is choice C, which adds the acceleration at t=2 to the initial velocity, confusing instantaneous acceleration with the accumulated change. In general, when solving initial value problems in kinematics, determine the constant of integration by applying the initial condition after finding the antiderivative.
A particle’s velocity is $v(t)=\ln(t+1)$ for $0\le t\le 3$ and $s(1)=5$; find $s(3)$.
$s(3)=5+\int_1^3\ln(t+1),dt$
$s(3)=5+\int_0^3\ln(t+1),dt$
$s(3)=5+\ln(4)$
$s(3)=\int_1^3\ln(t+1),dt$
$s(3)=5+\int_1^3\frac{1}{t+1},dt$
Explanation
This problem involves using integration in kinematics to connect velocity to position. To find the position at t=3, integrate the velocity function v(t)=ln(t+1) from 1 to 3, which gives the change in position over that interval. Then, add the initial position s(1)=5 to account for the starting point. This yields s(3)=5 + ∫ from 1 to 3 of ln(t+1) dt, with limits matching the given initial time. A tempting distractor is choice D, which adds the velocity at t=3 to the initial position, mistaking instantaneous velocity for displacement. In general, when solving initial value problems in kinematics, determine the constant of integration by applying the initial condition after finding the antiderivative.
A particle moves with velocity $v(t)=t^3-6t$ for $0 \le t \le 2$ and $s(0)=0$; find $s(2)$.
$s(2)=0+\int_0^2(t^3-6t),dt$
$s(2)=\int_0^2(t^3),dt$
$s(2)=\int_0^2(3t^2-6),dt$
$s(2)=\int_0^2(t^3-6t),dt$
$s(2)=t^3-6t\big|_{t=2}$
Explanation
This problem involves using integration in kinematics to connect velocity to position. To find the position at t=2, integrate the velocity function $v(t)=t^3-6t$ from 0 to 2, which gives the change in position over that interval. Then, add the initial position s(0)=0 to account for the starting point. This yields $s(2)=0 + \int_0^2 (t^3-6t) , dt$, as the integral computes the net displacement. A tempting distractor is choice C, which evaluates the velocity at t=2 without integrating, ignoring the accumulation over time. In general, when solving initial value problems in kinematics, determine the constant of integration by applying the initial condition after finding the antiderivative.
A particle has velocity $v(t)=3t^2-4t+1$ for $0\le t\le 3$ and position $s(0)=2$; find $s(3)$.
$s(3)=2+\int_0^3(3t^2-4t),dt$
$s(3)=2+(3(3)^2-4(3)+1)$
$s(3)=\int_0^3(3t^2-4t+1),dt$
$s(3)=2+\int_0^3(3t^2-4t+1),dt$
$s(3)=2+\int_0^3(6t-4),dt$
Explanation
This problem involves using integration in kinematics to connect velocity to position. To find the position at t=3, integrate the velocity function v(t)=3t²-4t+1 from 0 to 3, which gives the change in position over that interval. Then, add the initial position s(0)=2 to account for the starting point. This yields the expression s(3)=2 + ∫ from 0 to 3 of (3t²-4t+1) dt, as the integral computes the net displacement. A tempting distractor is choice C, which incorrectly evaluates the velocity function at t=3 and adds it to the initial position, mistaking instantaneous velocity for displacement. In general, when solving initial value problems in kinematics, determine the constant of integration by applying the initial condition after finding the antiderivative.
A particle has acceleration $a(t)=4\cos(2t)$ for $0\le t\le\frac{\pi}{2}$ with $v(0)=-1$; what is $v!\left(\frac{\pi}{2}\right)$?
-5
-3
-1
1
3
Explanation
This problem involves kinematics integration to find velocity from acceleration. Since acceleration a(t) = 4cos(2t) is the derivative of velocity, we integrate using the chain rule in reverse: v(t) = ∫4cos(2t)dt = 4·(1/2)sin(2t) + C = 2sin(2t) + C. Using the initial condition v(0) = -1, we find 2sin(0) + C = -1, so 0 + C = -1, giving C = -1. Therefore v(t) = 2sin(2t) - 1. Evaluating at t = π/2 gives v(π/2) = 2sin(π) - 1 = 2(0) - 1 = -1. A common error is forgetting to divide by the coefficient of t when integrating, which would incorrectly give 4sin(2t). Always account for the chain rule when integrating composite functions.