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 $$a(t)=\dfrac {dv}{dt}=\dfrac {d^2s}{dt^2}$$

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  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

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.

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 ? 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.

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

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  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.

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? 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