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