Velocity and acceleration

Differentiating position $x(t)$ once with respect to $t$ gives velocity $v$; differentiating twice gives acceleration $a$.

For example, if $x(t)=t^3-12t$, then $v=3t^2-12$ and $a=6t$. At $t=2$ (the positive root of $v=0$) the point changes direction. To find the velocity at a given time, find $v(t)$ first and then substitute the value of $t$.

Examples

1. 01What is the velocity at t=1?$\displaystyle x(t)=t^{2}-2t$AnswerHide

   Answer$\displaystyle 0$

   Steps

   1. First differentiate to get the velocity
   2. $\displaystyle v$$\displaystyle = 2t-2$
   3. Then put t = 1 in
   4. $\displaystyle 2 \cdot 1 -2$$\displaystyle = 0$
2. 02What is the velocity v?$\displaystyle x(t)=t^{2}-3t$AnswerHide

   Answer$\displaystyle 2t-3$

   Steps

   1. Differentiate term by term
   2. $\displaystyle t^{2}$$\displaystyle \to 2t$$\displaystyle -3t$$\displaystyle \to -3$
3. 03What is the acceleration a?$\displaystyle v(t)=9t^{2}-4$AnswerHide

   Answer$\displaystyle 18t$

   Steps

   1. Differentiate term by term
   2. $\displaystyle 9t^{2}$$\displaystyle \to 18t$$\displaystyle -4$$\displaystyle \to 0$
4. 04At what time t is the velocity 0? (the positive one)$\displaystyle x(t)=t^{3}-12t,\; v(t)=3t^{2}-12$AnswerHide

   Answer$\displaystyle 2$

   Steps

   1. $\displaystyle 3t^{2}-12 = 0$$\displaystyle \Rightarrow t^{2} = 4$
   2. Time, so take the positive root
   3. $\displaystyle t$$\displaystyle = 2$
5. 05What is the velocity v?$\displaystyle x(t)=t^{2}-5t$AnswerHide

   Answer$\displaystyle 2t-5$

   Steps

   1. Differentiate term by term
   2. $\displaystyle t^{2}$$\displaystyle \to 2t$$\displaystyle -5t$$\displaystyle \to -5$
6. 06What is the acceleration a?$\displaystyle v(t)=2t-2$AnswerHide

   Answer$\displaystyle 2$

   Steps

   1. Differentiate term by term
   2. $\displaystyle 2t$$\displaystyle \to 2$$\displaystyle -2$$\displaystyle \to 0$

FAQ

Q1What is the difference between velocity and speed?AnswerHide

A

Velocity includes direction and can be negative. Speed is its absolute value $|v|$. If $v=-3$, the point moves in the negative direction at speed 3.

Q2If acceleration is positive, is the point speeding up?AnswerHide

A

Not necessarily. The speed increases when $v$ and $a$ have the same sign. If $v<0$ and $a>0$, the point is slowing down.