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Application Of Rate Of Change In Physics

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

Rate of Change in Physics

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.

Illustration Questions

A particle moves in a straight line with position function s(t) = t3 –4t2 + 4 t where 't' is in seconds and 's' in meters. The velocity of the particle after 4 seconds.

A 52 m/sec.

B –16 m/sec.

C 10 m/sec.

D 20 m/sec.

×

\(v(t)=\dfrac {ds}{dt}=\dfrac {d}{dt}(t^3-4t^2+4t)\)

\(=3t^2-8t+4\)

\(\therefore \;v(t)=3t^2-8t+4\)

at t=4,

= 3 × 42 – 8 × 4 + 4 

= 3 × 16 – 8 × 4 + 4

= 20 m/sec.

A particle moves in a straight line with position function s(t) = t3 –4t2 + 4 t where 't' is in seconds and 's' in meters. The velocity of the particle after 4 seconds.

A

52 m/sec.

.

B

–16 m/sec.

C

10 m/sec.

D

20 m/sec.

Option D is Correct

Condition for Particle to move in Forward Direction

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

Illustration Questions

A particle moves in a straight line with position function s(t) = t3 –10t2 + 25t where,  't' is in seconds and 's' in meters. When is the particle moving forwards?

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

×

Particle moves in forward direction when, v(t) > 0

\(v(t)=\dfrac {ds}{dt}\)

\(=\dfrac {d}{dt}(t^3-10t^2+25t)\)

= 3t2 –20t + 25 

v(t) > 0,

 \(\Rightarrow\) 3t2 –20t + 25 > 0

\(\Rightarrow\) (t –5) (3t –5) > 0

\(\Rightarrow\) t > 5 or t < 5/3 (both factor positive or both negative)

\(\therefore\) Required time interval t > 5 or t < 5/3

A particle moves in a straight line with position function s(t) = t3 –10t2 + 25t where,  't' is in seconds and 's' in meters. When is the particle moving forwards?

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

Option A is Correct

Condition for Particle to move in Backward Direction

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.

Illustration Questions

A particle moves in a straight line with position function s(t) = t3 –10t2 + 25t where  't' is in seconds and 's' in meters; when is the particle moving backward?

A 6 < t < 7

B 1/5 < t < 2

C 5/3 < t < 5

D 2 < t < 4

×

Particle moves in the backward direction when v(t) < 0

\(v(t)=\dfrac {ds}{dt}\)

\(=\dfrac {d}{dt}(t^3-10t^2+25t)\)

= 3t2 –20t + 25 

v(t) < 0 \(\Rightarrow\) 3t2 –20t + 25 < 0

\(\Rightarrow\) (t –5) (3t –5) < 0

\(\Rightarrow\)  5/3 < t < 5 

\(\therefore\) Required time interval 5/3 < t < 5 

A particle moves in a straight line with position function s(t) = t3 –10t2 + 25t where  't' is in seconds and 's' in meters; when is the particle moving backward?

A

6 < t < 7

.

B

1/5 < t < 2

C

5/3 < t < 5

D

2 < t < 4

Option C is Correct

Finding the Total Distance Traveled for the given Displacement Function

  • 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.

Illustration Questions

A particle moves in a straight line with position function s(t) = t3 –9t2 + 24t, where 't' is in seconds and 's' in meters. Find the total distance traveled during first 6 seconds.

A 55 m

B 44 m

C 716 m

D 27 m

×

First we calculate the time interval in which particle moves forward & backward i.e.

v(t) > 0 or < 0

v(t) = \(\dfrac {ds}{dt}=\dfrac {d}{dt}\)(t3 –9t2 + 24t)

= 3t2 – 18t + 24

v(t) > 0 

\(\Rightarrow\)  3 (t2 –6t + 8) > 0 

\(\Rightarrow\) (t –4) (t –2) > 0

\(\Rightarrow\) t > 4 or t < 2

v(t) < 0 

\(\Rightarrow\)  3 (t2 –6t + 8) < 0 

\(\Rightarrow\) 3(t –4) (t –2) < 0

\(\Rightarrow\) 2 < t < 4 

Distance traveled during 6 seconds =

= |s(2) – s(0)| + |s(4) –s(2)| + |s(6) – s(4)|

= |(20–0)| + |(16–20)| + |(36–16)|

= 20 + 4 + 20

= 44 m

A particle moves in a straight line with position function s(t) = t3 –9t2 + 24t, where 't' is in seconds and 's' in meters. Find the total distance traveled during first 6 seconds.

A

55 m

.

B

44 m

C

716 m

D

27 m

Option B is Correct

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