Connecting Position, Velocity, and Acceleration - AP Calculus BC

A motorcycle's acceleration is $a(t)=2\cos t$ with $v(0)=-1$ ; what is $v\left(\frac{\pi}{2}\right)$ ?
Answer: $v\left(\frac{\pi}{2}\right)=-1+\int_0^{\pi/2}2\cos t\,dt$ 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 particle has velocity $v(t)=3t^2-4t+1$ (m/s) and position $s(0)=5$ m; find $s(2)$ .
Answer: $5+\int_{0}^{2}(3t^2-4t+1)\,dt$ This problem requires using kinematics integration to find position from velocity. Since velocity v(t) = ds/dt, we have s(t) = s(0) + ∫₀ᵗ v(u)du by the Fundamental Theorem of Calculus. Given s(0) = 5 and v(t) = 3t² - 4t + 1, we get s(2) = 5 + ∫₀² (3t² - 4t + 1)dt. Choice D incorrectly attempts to evaluate the velocity function at the endpoints rather than integrating it. When connecting velocity to position, always integrate the velocity function and add the initial position.
A cyclist's acceleration is $a(t)=t^2-1$ with $v(2)=0$ ; what is $v(6)$ ?
Answer: $0+\int_{2}^{6}(t^2-1)\,dt$ This problem involves kinematics integration to find velocity from acceleration. The velocity satisfies v(t) = v(t₀) + ∫_{t₀}ᵗ a(u)du. Given v(2) = 0 and a(t) = t² - 1, we have v(6) = 0 + ∫₂⁶ (t² - 1) dt. Choice E incorrectly uses 2t, which is the derivative of t² rather than the acceleration function itself. Always integrate the given acceleration function from the initial time to the final time.
A car's acceleration is $a(t)=6t-2$ and $v(0)=4$ ; what is the velocity $v(3)$ ?
Answer: $4+\int_0^3 (6t-2)\,dt$ This problem involves kinematics integration to find velocity from acceleration. Since acceleration a(t) = 6t - 2 is the derivative of velocity, we integrate to find velocity: v(t) = v(0) + ∫₀ᵗ a(u)du. Given v(0) = 4, we have v(3) = 4 + ∫₀³ (6t - 2)dt. Choice D incorrectly adds the acceleration value at t = 3 instead of integrating over the entire interval, missing the cumulative effect of acceleration. To find velocity change, always integrate acceleration over the time interval and add the initial velocity.
A particle's velocity is $v(t)=\sqrt{t}$ for $0\le t\le4$ and $s(0)=-2$ ; what is $s(4)$ ?
Answer: $\frac{10}{3}$ This problem requires kinematics integration to find position from velocity. Since velocity v(t) = √t = $t^{1/2}$ is the derivative of position, we integrate using the power rule: s(t) = ∫ $t^{1/2}$ dt = $t^{3/2}$ /(3/2) + C = (2/3) $t^{3/2}$ + C. Using the initial condition s(0) = -2, we find (2/3)(0) + C = -2, so C = -2. Therefore s(t) = (2/3) $t^{3/2}$ - 2. Evaluating at t = 4 gives s(4) = (2/3) $(4)^{3/2}$ - 2 = (2/3)(8) - 2 = 16/3 - 2 = 16/3 - 6/3 = 10/3. A common error is computing $4^{3/2}$ as 6 instead of 8, forgetting that $4^{3/2}$ = (√4)³ = 2³ = 8. When dealing with fractional exponents, convert to radical form if needed for clarity.
A particle's velocity is $v(t)=\frac{1}{1+t^2}$ for $0\le t\le 2$ and $s(0)=7$ ; find $s(2)$ .
Answer: $s(2)=7+\int_0^2\frac{1}{1+t^2}\,dt$ 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)=1/(1+t²) from 0 to 2, which gives the change in position over that interval. Then, add the initial position s(0)=7 to account for the starting point. This yields s(2)=7 + ∫ from 0 to 2 of 1/(1+t²) dt, as the integral computes the net displacement. A tempting distractor is choice C, which adds the velocity at t=2 to the initial position, confusing instantaneous velocity with accumulated displacement. In general, when solving initial value problems in kinematics, determine the constant of integration by applying the initial condition after finding the antiderivative.
An object's velocity is $v(t)=\frac{2}{1+t^2}$ (m/s) and $s(0)=-3$ m; find $s(1)$ .
Answer: $-3+\int_{0}^{1}\frac{2}{1+t^2}\,dt$ This question requires kinematics integration to find position from velocity. Using $s(t) = s(0) + \int_0^t v(u)\, du$ , with $s(0) = -3$ and $v(t) = \frac{2}{1+t^2}$ , we get $s(1) = -3 + \int_0^1 \frac{2}{1+t^2}\, dt$ . Choice E incorrectly shows the derivative of the velocity function rather than integrating the velocity itself. When finding position from velocity, integrate the velocity function and add the initial position value.
A particle has velocity $v(t)=3t^2-4t+1$ for $0\le t\le2$ and position $s(0)=5$ ; find $s(2)$ .
Answer: $5+\int_0^2 (3t^2-4t+1)\,dt$ 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 car's acceleration is $a(t)=4t-6$ (m/s $^2$ ) with $v(0)=5$ ; what is $v(2)$ ?
Answer: $v(2)=5+\int_0^2(4t-6)\,dt$ 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 has velocity $v(t)=\sin t+t$ for $0\le t\le \pi$ and $s(0)=-1$ ; find $s(\pi)$ .
Answer: $s(\pi)=-1+\int_0^{\pi}(\sin t+t)\,dt$ 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 cyclist's acceleration is $a(t)=t-\frac{2}{t+1}$ with $v(0)=0$ ; what is $v(3)$ ?
Answer: $v(3)=0+\int_0^3\left(t-\frac{2}{t+1}\right)\,dt$ This problem involves using integration in kinematics to connect acceleration to velocity. To find the velocity at t=3, integrate the acceleration function a(t)=t - 2/(t+1) from 0 to 3, which gives the change in velocity over that interval. Then, add the initial velocity v(0)=0 to obtain the final velocity. This results in v(3)=0 + ∫ from 0 to 3 of (t - 2/(t+1)) dt, where the integral represents the net change due to acceleration. A tempting distractor is choice C, which evaluates the acceleration at t=3 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 acceleration $a(t)=4\cos(2t)$ and $v(0)=3$ ; what is $v\!\left(\tfrac{\pi}{2}\right)$ ?
Answer: $3+\int_0^{\pi/2} 4\cos(2t)\,dt$ This problem involves kinematics integration to find velocity from acceleration. Since acceleration a(t) = 4cos(2t) is the derivative of velocity, we integrate: v(t) = v(0) + ∫₀ᵗ a(u)du. Given v(0) = 3, we have v(π/2) = 3 + ∫₀^(π/2) 4cos(2t)dt. Choice D incorrectly uses -8sin(2t) as the integrand, which is the derivative of 4cos(2t) rather than the acceleration itself. To find velocity from acceleration, integrate the given acceleration function, not its derivative.
A particle's acceleration is $a(t)=2t+\sin t$ and $v(0)=1$ ; what is $v(\pi)$ ?
Answer: $1+\int_0^{\pi} (2t+\sin t)\,dt$ This problem requires kinematics integration to find velocity from acceleration. Since acceleration a(t) = 2t + sin t is the derivative of velocity, we integrate: v(t) = v(0) + ∫₀ᵗ a(u)du. Given v(0) = 1, we have v(π) = 1 + ∫₀ᵖ (2t + sin t)dt. Choice D incorrectly uses 2 + cos t as the integrand, which would be the derivative of 2t + sin t rather than the acceleration itself. To find velocity change, always integrate the given acceleration function over the time interval.
A particle's velocity is $v(t)=\ln(t+1)$ for $0\le t\le 3$ and $s(1)=5$ ; find $s(3)$ .
Answer: $s(3)=5+\int_1^3\ln(t+1)\,dt$ 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.
An object's acceleration is $a(t)=\frac{1}{t+1}$ with $v(0)=-1$ ; find $v(3)$ .
Answer: $-1+\int_{0}^{3}\frac{1}{t+1}\,dt$ This problem requires kinematics integration to determine velocity from acceleration. The velocity follows $v(t) = v(0) + \int_0^t a(u) \, du$ . With $v(0) = -1$ and $a(t) = \frac{1}{t+1}$ , we obtain $v(3) = -1 + \int_0^3 \frac{1}{t+1} \, dt$ . Choice E incorrectly uses $\frac{-1}{(t+1)^2}$ , which is the derivative of $\frac{1}{t+1}$ rather than the acceleration function itself. When finding velocity from acceleration, integrate the acceleration function and add the initial velocity.
A particle moves with velocity $v(t)=t^3-6t$ for $0 \le t \le 2$ and $s(0)=0$ ; find $s(2)$ .
Answer: $s(2)=0+\int_0^2(t^3-6t)\,dt$ 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's acceleration is $a(t)=e^t$ and $v(0)=-1$ ; determine $v(1)$ .
Answer: $-1+\int_0^1 e^t\,dt$ This problem requires kinematics integration to find velocity from acceleration. Since acceleration a(t) = eᵗ is the derivative of velocity, we integrate: v(t) = v(0) + ∫₀ᵗ a(u)du. Given v(0) = -1, we have v(1) = -1 + ∫₀¹ eᵗ dt. Choice A incorrectly evaluates this as -1 + e, which assumes ∫₀¹ eᵗ dt = e rather than e - 1. Remember that when integrating exponentials, ∫₀¹ eᵗ dt = eᵗ|₀¹ = e¹ - e⁰ = e - 1.
A particle's velocity is $v(t)=\frac{2}{1+t^2}$ for $0\le t\le1$ and $s(0)=0$ ; what is $s(1)$ ?
Answer: $2\arctan(1)$ This problem requires kinematics integration to find position from velocity. Since velocity v(t) = 2/(1+t²) is the derivative of position, we integrate using the arctangent formula: s(t) = ∫2/(1+t²)dt = 2arctan(t) + C. Using the initial condition s(0) = 0, we find 2arctan(0) + C = 0, so 0 + C = 0, giving C = 0. Therefore s(t) = 2arctan(t). Evaluating at t = 1 gives s(1) = 2arctan(1) = 2(π/4) = π/2. A common mistake is forgetting the coefficient 2 when integrating, which would give just arctan(1). Remember that constants factor out of integrals and must be preserved throughout the calculation.
A car's acceleration is $a(t)=2t-3$ (m/s $^2$ ) with $v(0)=4$ m/s; what is $v(5)$ ?
Answer: $4+\int_{0}^{5}(2t-3)\,dt$ This problem involves kinematics integration to find velocity from acceleration. Since acceleration a(t) = dv/dt, we have v(t) = v(0) + ∫₀ᵗ a(u)du. With v(0) = 4 and a(t) = 2t - 3, we obtain v(5) = 4 + ∫₀⁵ (2t - 3)dt. Choice E incorrectly integrates only the constant term from the derivative of acceleration, confusing the relationship between a(t) and v(t). To find velocity from acceleration, integrate the acceleration function and add the initial velocity.
An object's velocity is $v(t)=t e^{-t}$ and $s(0)=10$ ; find the position $s(2)$ .
Answer: $10+\int_0^2 t e^{-t}\,dt$ This problem requires kinematics integration to find position from velocity. Since velocity v(t) = te⁻ᵗ is the derivative of position, we integrate: s(t) = s(0) + ∫₀ᵗ v(u)du. With s(0) = 10, we get s(2) = 10 + ∫₀² te⁻ᵗ dt. Choice E incorrectly uses e⁻ᵗ - te⁻ᵗ as the integrand, which appears to be an attempt at the derivative rather than the velocity itself. When finding position, always integrate the given velocity function directly.

Start this quiz