Informative line

Rate Of Change Of Velocity

Learn velocity and acceleration formula in calculus, practice for finding the acceleration & speeding up and slowing down interval from velocity graph.

Finding the acceleration

Acceleration is given by \(a(t)=\dfrac {dv}{dt}=\dfrac {d^2s}{dt^2}\)

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. What is the value of acceleration at t = 3 seconds?

A 0 m/s2 \(\)

B 5 m/s2

C –7 m/s2

D 18 m/s2

×

\(a(t)=\dfrac {d^2s}{dt^2}=\dfrac {d}{dt} \left ( \dfrac {ds}{dt} \right)\)

\(=\dfrac {d}{dt} \left ( \dfrac {d}{dt} (t^3-9t^2+24t) \right)\)

\(=\dfrac {d}{dt} \left ( 3t^2-18t+24 \right)\)

\(= 6 t – 18\)

\(a(t) = 6t – 18 \)

\(\Rightarrow\)\(a(3) = 18 – 18 = 0\,m/s^2\)

A particle moves in a straight line with position function s(t) = t3 –9t2 + 24t where 't' is in seconds and 's' in meters. What is the value of acceleration at t = 3 seconds?

A

0 m/s2 \(\)

.

B

5 m/s2

C

–7 m/s2

D

18 m/s2

Option A is Correct

Speeding Up of a Particle 

  • A particle speeds up when velocity and acceleration both are of same sign i.e. either both are positive or both are negative.
  • Speeding up \(\Rightarrow\) a > 0 or v > 0 or a < 0 and v < 0

Illustration Questions

A particle moves according to the law of motion s(t) = 2t3 –27t2 + 84t where 't' is in seconds and 's' is in meters. When is the particle speeding up?

A t > 7 and 2 < t < 4.5

B t > 9

C t > 8 or 3 < t < 5

D t < 3

×

Speeding up \(\Rightarrow\)     a > 0 & v > 0 or a < 0 & v < 0

v(t) = \(\dfrac {ds}{dt}=\) 6t2–54t+84  = 6(t2–9t+14)

a(t) = \(\dfrac {d^2s}{dt^2}=\) \(\dfrac {d}{dt}\)6(t2 – 9t +14)

= 6 (2t – 9) 

Case 1 :

a > 0

 \(\Rightarrow\) 6 (2t – 9) > 0  

     \(\Rightarrow t>\dfrac {9}{2}\)

v > 0

 \(\Rightarrow\) 6(t– 9t + 14) > 0    

  \(\Rightarrow\) 6(t –7) (t –2) > 0

\(\Rightarrow\) t > 7 or t < 2

\(\therefore\) a > 0 and v > 0

 \(\Rightarrow\) t > 7

Case 2 : 

a < 0  

     \(\Rightarrow\) 6 (2t – 9) < 0     \(\Rightarrow t<\dfrac {9}{2}\)

v < 0    

  \(\Rightarrow\) 6(t– 9t + 14) < 0    

  \(\Rightarrow\) 6(t –7) (t –2) < 0

\(\;\;\Rightarrow\) 2 < t < 7

\(\therefore\)  a < 0 and v < 0

 \(\Rightarrow\) 2 < t < 4.5

\(\therefore\) Speeding up intervals \(\rightarrow\) t > 7 and 2 < t < 4.5

A particle moves according to the law of motion s(t) = 2t3 –27t2 + 84t where 't' is in seconds and 's' is in meters. When is the particle speeding up?

A

t > 7 and 2 < t < 4.5

.

B

t > 9

C

t > 8 or 3 < t < 5

D

t < 3

Option A is Correct

Slowing Down of a Particle

  • A particle slows down when velocity and acceleration both are of opposite signs i.e. a > 0 and v < 0 or a < 0 and v > 0.
  • Slowing down   \(\Rightarrow\) a > 0 and v < 0 or a < 0 & v > 0 

Illustration Questions

A particle moves according to the law of motion s(t) = 2t3 –27t2 + 84t where 't' is in seconds and 's' is in meters. When is the particle slowing down?

A 9/2 < t < 7 and t < 2

B t > 9

C 2 < t < 5

D 3/2 < t < 5 or t < 1

×

Slowing down   \(\Rightarrow\) a > 0 and v < 0 or a < 0 and v > 0

v(t) = \(\dfrac {ds}{dt}=\) 6t2 –54t  + 84  

= 6(t2 – 9t +14)

a(t) = \(\dfrac {d^2s}{dt^2}=\)\(\dfrac {d}{dt}\) 6(t2 – 9t +14)

= 6 (2t – 9) 

Case 1 :

a > 0

 \(\Rightarrow\) 6 (2t – 9) > 0

 \(\Rightarrow t>\dfrac {9}{2}\)

v < 0

 \(\Rightarrow\) 6(t– 9t + 14) < 0 

\(\Rightarrow\) 6(t –7) (t –2) < 0

\(\Rightarrow\) 2< t <7

\(\therefore\) a > 0 and v < 0

 \(\Rightarrow\) 9/2 < t < 7

Case 2 : 

a < 0

 \(\Rightarrow\) 6 (2t – 9) < 0

\(\Rightarrow t<\dfrac {9}{2}\)

v > 0

 \(\Rightarrow\)6(t– 9t + 14) > 0

 \(\Rightarrow\) 6(t –7) (t –2) >0

\(\Rightarrow\)  t > 7 or t < 2

\(\therefore\) a < 0 and v > 0

 \(\Rightarrow\)  t < 2

\(\therefore\) Slowing down interval is  \(\rightarrow\) 9/2 < t < 7  or t < 2

A particle moves according to the law of motion s(t) = 2t3 –27t2 + 84t where 't' is in seconds and 's' is in meters. When is the particle slowing down?

A

9/2 < t < 7 and t < 2

.

B

t > 9

C

2 < t < 5

D

3/2 < t < 5 or t < 1

Option A is Correct

Identifying the Speeding up Interval from Velocity-time Graph

  • A particle is said to speed up under two conditions:
  1. If velocity and acceleration are both positive.
  2. If velocity and acceleration are both negative.

i.e., If a > 0 & v > 0 or a < 0 & v < 0 

where, a = acceleration and v = velocity

  • If velocity - time graph is given then time of speeding up is the part of graph when it is above x-axis and increasing or below x axis and decreasing.

Illustration Questions

Given is the graph of velocity function v(t) v/s time. When is the particle speeding up?

A t > 4 and 5 < t < 6

B 5 < t < 7

C 2 < t < 5 and 0 < t < 1

D 0 < t < 6

×

Speeding up

  \(\Rightarrow\) a > 0 &  v > 0 or a < 0 & v < 0

The velocity graph should be above the x-axis (v > 0) and should be increasing \(\left ( \dfrac {dv}{dt}=a>0 \right)\)

or below the x-axis (v < 0) and decreasing (a < 0 )

\(\therefore\) Required interval is \(\rightarrow\) 2 < t < 5 or 0 < t < 1

Given is the graph of velocity function v(t) v/s time. When is the particle speeding up?

image
A

t > 4 and 5 < t < 6

.

B

5 < t < 7

C

2 < t < 5 and 0 < t < 1

D

0 < t < 6

Option C is Correct

Identifying the Slowing down Interval from Velocity-time Graph

  • A particle is said to be slowing down under two conditions:
  1. If velocity is positive and acceleration is negative.
  2. If velocity is negative and acceleration is positive.

i.e., If a > 0 & v < 0 or a < 0 & v > 0 

where, a = acceleration and v = velocity

  • If velocity - time graph is given then time of slowing down is the part of graph when it is above the x-axis and decreasing or below x axis and increasing.

Illustration Questions

Given is the graph of velocity function v(t) v/s time. When is the particle slowing down ?

A 2 < t < 4 or 6 < t < 7

B t > 5

C 1 < t < 2 or 5 < t < 7

D t < 4

×

Slowing down 

  \(\Rightarrow\) a < 0 & v > 0  or  a > 0 & v < 0

The velocity graph should be above the \(x\) axis (v > 0) and decreasing (a < 0) or below the \(x\)axis v < 0 and increasing ( a > 0 ).

\(\therefore\) Required interval is \(\rightarrow\) 1 < t < 2 or 5 < t < 7

Given is the graph of velocity function v(t) v/s time. When is the particle slowing down ?

image
A

2 < t < 4 or 6 < t < 7

.

B

t > 5

C

1 < t < 2 or 5 < t < 7

D

t < 4

Option C is Correct

Problems on Motion of Particle Thrown Vertically Upwards

  • When a particle is thrown vertically upwards if reaches a particular height and then comes down due to the effect of gravity, 
  • To find the maximum height reached by particle first fid the time at which the velocity is 0 i.e. particle is momentarily at rest then find height by putting the value.  

Illustration Questions

A ball is thrown vertically upwards with a velocity of \(20\; m/s\).Its height after 't' seconds is given by \(s=20\; t - 4 t^2 \) where 's' is in meters. What is the max height reached by the ball ?

A 56 m 

B 25 m

C 108 m

D 2 m

×

Max height will occur where velocity is 0.

i.e particle starts returning down.

\(V = \dfrac{ds}{dt} = \dfrac{d}{dt }\; ( 20 \; t - 4 \; t ^2) = 20 - 8t = 0\)

\(\Rightarrow\;\; t = \dfrac{5}{2}\; s\)

\(\therefore\;\; v = 0 \) when \( t = \dfrac{5}{2}\; s \)

\(\Rightarrow\) maximum height \( = 20 × \dfrac{5}{2} - 4 × \dfrac{25}{4}\)

\( = 50 - 25 = 25 \; m\)

A ball is thrown vertically upwards with a velocity of \(20\; m/s\).Its height after 't' seconds is given by \(s=20\; t - 4 t^2 \) where 's' is in meters. What is the max height reached by the ball ?

A

56 m 

.

B

25 m

C

108 m

D

2 m

Option B is Correct

Practice Now