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Calculus IApplications

Motion and antiderivatives, forwards and backwards.

Read motion from a derivative, then undo derivatives to recover a function. Explore velocity, distance and the constant C before solving class questions and extra practice.

Motion and rates · Step 1 of 9

Position says where; velocity says how it is changing

Horizontal: time in seconds. Vertical: position in metres. The dot follows your chosen time.

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Position s(t){s(t)} is a signed location on a line. Velocity is its derivative: v(t)=s′(t){v(t)} = {s'(t)}. On a position graph, velocity is the slope, not the height. For this journey, position is in metres and time is in seconds.
Try it: Move the time slider along s(t)=t2−4t+3{s(t)} = {t^2-4t+3}. At time 1, the position is zero. Is the particle stopped there?
Time (seconds)1
v(t)=2t−4{\displaystyle v(t)} = {\displaystyle 2t} - {\displaystyle 4}
s(1)=(1)2−4(1)+3=0{\displaystyle s(1)} = {\displaystyle (1)^2} - {\displaystyle 4(1)} + {\displaystyle 3} = {\displaystyle 0}
v(1)=2(1)−4=−2{\displaystyle v(1)} = {\displaystyle 2(1)} - {\displaystyle 4} = {\displaystyle -2}

A zero position means passing the origin. A zero velocity means being at rest at that instant.

vaverage=s(t2)−s(t1)t2−t1{\displaystyle v_{\rm average}} = {\displaystyle \frac{s(t_2)-s(t_1)}{t_2-t_1}}

Average velocity compares two times; the derivative gives velocity at one instant.

Check yourself
The position graph is above zero but falling. Which way is the particle moving?

Need a hand with motion or finding the original function?

Bring your practice questions. We work through them together until every type feels routine.

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