Learn rate of change calculus and physics problems, instantaneous velocity in calculus. Practice for finding the total distance traveled calculus given the displacement function.

If s = f(t) is the position of a particle that is moving in a straight line, then

\(\dfrac {\Delta s}{\Delta t}=\)average velocity of particle over time \(\Delta t\)

\(\dfrac {ds}{dt}=\)instantaneous velocity (rate of change of displacement w.r.t. time)

= v(t)

\(\dfrac {dv}{dt}=\dfrac {d^2 s}{dt}\) = instantaneous acceleration (rate of change of velocity w.r.t. time)

Slope of chord \(PQ=\dfrac {s(t_2)-s(t_1)}{t_2-t_1}\)

= average velocity

\(\dfrac {ds}{dt}=\) slope of tangent at P

= instantaneous velocity at P.

A 52 m/sec.

B –16 m/sec.

C 10 m/sec.

D 20 m/sec.

We say that a particle is moving forward if its instantaneous velocity is positive while it moves backwards if instantaneous velocity is negative.

i.e.,

\(\dfrac {ds}{dt}=v>0\rightarrow\) particle moves forward

A t > 5 or t < 5/3

B t > 3/5 or t < 1

C t > 4 or t < 1

D t > 5 or t < 2

We say that a particle is moving forward if its instantaneous velocity is positive while it moves backwards if instantaneous velocity is negative.

i.e.,

\(\dfrac {ds}{dt}=v<0\rightarrow\) particle moves backward.

A 6 < t < 7

B 1/5 < t < 2

C 5/3 < t < 5

D 2 < t < 4

- Consider the path of a particle (Shown by an arrow)

Total displacement after 10 seconds = 5 m

Total distance after 10 seconds = 7 + 3 + 1 = 11 m

- Distance traveled \(=\underbrace{\Big(s(5)-s(0)\Big)}_{\text {forward journey}}+\underbrace {|s(8)-s(5)|}_{\text {backward journey}}+\underbrace{\Big(s(10)-s(8)\Big)}_{\text {forward journey}}\)

\(= ( 7 – 0 ) + | 4 –7 | + (5 – 4)\)

\(= 7 + 3 + 1 = 11\)

**Steps to find distance traveled:**

(1) Find \(v= \dfrac {ds}{dt}\)

(2) Put v = 0, then find values of 't' for which v > 0 or v < 0.

(3) Then find the distance traveled in separate intervals and add.