AP Calculus AB and BC: Chapter 3 – Applications of Differentiation :3.2 -Position, Velocity, and Acceleration Study Notes

Position, Velocity, and Acceleration

Let $s(t)$ be the position function for an object moving along a straight line. Velocity is the derivative of position with respect to time.
$
v(t)=\lim _{h \rightarrow 0} \frac{s(t+h)-s(t)}{h}=s^{\prime}(t) .
$
Speed is the absolute value of velocity.
$
\text { speed }=|v(t)|
$
Acceleration is the derivative of velocity with respect to time.
$
a(t)=v^{\prime}(t)=s^{\prime \prime}(t)
$

If $v>0$, then the particle is moving to the right.
If $v<0$, then the particle is moving to the left.
If $a>0$, then $v$ is increasing.
If $a<0$, then $v$ is decreasing.
If $a$ and $v$ have the same sign, the particle’s speed is increasing.
If $a$ and $v$ have the opposite signs, the particle’s speed is decreasing.
Note: For movement on a horizontal line, $x(t)$ is used to represent the position function; rightward movement is considered to be in the positive direction.
For movement on a vertical line, $y(t)$ is used to represent the position function; upward movement is considered to be in the positive direction.

Example 

  • A particle starts moving at time $t=0$ and moves along the $x$-axis so that its position at time $t \geq 0$ is given by $x(t)=t^3-\frac{9}{2} t^2+7$.

(a) Find the velocity of the particle at any time $t \geq 0$.
(b) For what values of $t$ is the particle moving to the left.
(c) Find the values of $t$ for which the particle is moving but its acceleration is zero.
(d) For what values of $t$ is the speed of the particle decreasing?

▶️Answer/Explanation

Solution
(a) $v(t)=x^{\prime}(t)=3 t^2-9 t$
(b) The particle is moving to the left when $v(t)<0$. $v(t)=3 t^2-9 t=3 t(t-3)<0 \Rightarrow 0<t<3$.
(c) $a(t)=v^{\prime}(t)=6 t-9=3(2 t-3)$
The acceleration is zero when $2 t-3=0$. So, $t=3 / 2$.

(d) The speed of particle is decreasing when $\frac{3}{2}<t<3$, since within this interval, $a(t)$ and $v(t)$ have the opposite signs.

Exercises – Position, Velocity, and Acceleration

Multiple Choice Questions

  • A particle moves along the $x$-axis so that at any time $t \geq 0$, its position is given by $x(t)=-\frac{1}{2} \cos t-3 t$. What is the acceleration of the particle when $t=\frac{\pi}{3}$ ?

(A) $-\frac{\sqrt{3}}{4}$

(B) $-\frac{1}{4}$

(C) $\frac{1}{4}$

(D) $\frac{\sqrt{3}}{4}$

▶️Answer/Explanation

Ans:C

Question

  •  A point moves along the $x$-axis so that at any time $t$, its position is given by $x(t)=\sqrt{x} \ln x$. For what values of $t$ is the particle at rest?

(A) No values

(B) $\frac{1}{e^2}$

(C) $\frac{1}{e}$

(D) $e$

▶️Answer/Explanation

Ans:B

Question

  • A particle moves along the $x$-axis so that at any time $t$, its position is given by $x(t)=3 \sin t+t^2+7$. What is velocity of the particle when its acceleration is zero?

(A) 1.504

(B) 1.847

(C) 2.965

(D) 3.696

▶️Answer/Explanation

Ans:D

Question

  •  Two particles start at the origin and move along the $x$-axis. For $0 \leq t \leq 8$, their respective position functions are given by $x_1(t)=\sin ^2 t$ and $x_2(t)=e^{-t}$. For how many values of $t$ do the particles have the same velocity?

(A) 3

(B) 4

(C) 5

(D) 6

▶️Answer/Explanation

Ans:C

Question

  • A particle moves along a line so that at time $t$, where $0 \leq t \leq 5$, its velocity is given by $v(t)=-t^3+6 t^2-15 t+10$. What is the minimum acceleration of the particle on the interval?

(A) -30

(B) -15

(C) -3

(D) 0

▶️Answer/Explanation

Ans:C

Question

  • A particle moves along the $x$-axis so that at any time $t \geq 0$, its velocity is given by $v(t)=-t^3 e^{-t}$. At what value of $t$ does $v$ attain its minimum?

(A) $3 \sqrt[3]{e}$

(B) 3

(C) 0

(D) $\sqrt[3]{e}$

▶️Answer/Explanation

Ans:A

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