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Position - Velocity - Acceleration Functions Worksheet

Position - Velocity - Acceleration Functions Worksheet
  • Page 1
 1.  
If the square of the displacement S traversed by a particle in time t is given by S2 = t2 + 9t + 64, then choose the acceleration of the particle from the following.
a.
1754s3
b.
175s2
c.
1754s4
d.
175s3


Solution:

Let S2 = t2 + 9t + 64

ddt(S2)=ddt(t2+9t+64)
[Take derivative of S.]

2SdSdt = 2t + 9

SV = t + 9 / 2
[As dsdt = velocity V.]

V = t+92s

dVdt=ddt(t+92S)
[Take derivative of V.]

= Sddt(t+92)-(t+92)dSdtS2
[Quotient rule of derivative.]

a = S-(t+92)VS2
[Acceleration = a = dvdt.]

= S-(t+92)2SS2
[Substitute V = t+92S.]

s²-t²-9t-814s3

t²+9t+64-t²-9t-814s3
[Substitute s² = t²+9t+64

= 4(64)-814s3 = 1754s3
[Simplify.]


Correct answer : (1)
 2.  
The displacement function of a particle with time t in seconds is S = 4t3 - 9t2 + 4t + 6 ft. Find the velocity of the particle when its acceleration becomes zero.
a.
4 ft/s
b.
54 ft/s
c.
6 ft/s
d.
-11 4 ft/s


Solution:

Here the displacement S = 4t3 - 9t2 + 4t + 4
[Definition.]

The velocity = V = dSdt

= ddt(4t3 - 9t2 + 4t + 6)
[Substitute S from step1.]

ÞV = 12t2 - 18t + 4
[Simplify.]

The acceleration = a = dVdt
[Definition.]

= ddt (12t2 - 18t + 4)
[Substitute V from step 4.]

Þ a = 24t - 18

Suppose a = 0 then 24t - 18 = 0 ® t = 3 / 4

So, the acceleration of the particle is zero, at t = 3 / 4sec.

At t = 3 / 4, the velocity of the particle = 12(3 / 4)2 - 18(3 / 4) + 4 = -11 / 4 ft/s.


Correct answer : (4)
 3.  
The displacement S of a particle with respect to time t is given by S = - 8t3 + 4t. The velocity of the particle
a.
Decreases with time
b.
Is constant with respect to time
c.
Increases and decreases with time
d.
Increases with time


Solution:

Here the displacement S = - 8t3 + 4t

The velocity of the particle V = dSdt
[Definition.]

= ddt (- 8t3 + 4t)
[Substitute S from step1.]

Þ V = - 24t2 + 4

The acceleration of the particle = a = dVdt
[Definition.]

= ddt (- 24t2 + 4)
[Substitute V from step4.]

Þ a = - 48t

The acceleration is negative for t > 0.

As acceleration is negative, with time t, the velocity of the particle decreases with time t.


Correct answer : (1)
 4.  
The distance 'S' on a straight line traveled by a particle in time t is given by S = 5t - 2t3. What is the maximum velocity of the particle?
a.
-1
b.
- 5
c.
5
d.
7


Solution:

Here the displacement S = 5t - 2t3
[The distance on a straight line is displacement.]

The velocity of the particle = V = dSdt
[Definitions.]

= ddt (5t - 2t3)
[Substitute S from step1.]

Þ V = 5 - 6t2

The maximum velocity of the particle = 5 - minimum value of 6t2

= 5 - 0
[Minimum value of t2 is 0.]

= 5


Correct answer : (3)
 5.  
The displacement 'S' described by a particle in t seconds is given by S = 2t - 2t2 ft . What is the velocity of the particle at any time t?
a.
(2 - 2t) ft/s
b.
2 ft/s
c.
- 4 ft/s
d.
(2 - 4t) ft/s


Solution:

Here displacement = S = 2t - 2t2 ft.

Velocity of the particle at any time t = dSdt
[Definition.]

= ddt (2t - 2t2)
[Substitute S from step1.]

= (2 - 4t) ft/s.


Correct answer : (4)
 6.  
The displacement S of a body in t seconds is given by S = (4t3 - 7t2 + 4t + 4) m. The instantaneous velocity of the body is
a.
(12t2 + 14t + 4) m/s
b.
(12t2 - 14t + 4) m/s
c.
(12t2 + 14t - 4) m/s
d.
( 12t2 + 14t) m/s


Solution:

Here the displacement = S = (4t3 - 7t2 + 4t + 4) m.

The instantaneous velocity of the particle = dSdt
[Definition.]

= ddt (4t3 - 7t2 + 4t + 4)
[Substitute S from step1.]

= (12t2 - 14t + 4) m/s.


Correct answer : (2)
 7.  
The displacement of a body with time t is related as S = ae6t - be-4t. Find the rate of change of displacement with respect to time t.
a.
6ae6t + 4be- 4t
b.
ae6t - be-4t
c.
ae6t + be-4t
d.
36ae6t - 16be-4t


Solution:

Here the displacement = S = ae6t - be- 4t

The rate of change of displacement with respect to time t = dSdt
[Definition.]

= ddt (ae6t - be- 4t)
[Substitute S from step1.]

= 6ae6t + 4be- 4t
[Simplify.]


Correct answer : (1)
 8.  
The displacement S (in meters) of a particle is related with time t(in seconds) as S = 3t3 + 4t2 + 3t + 2 The velocity of the particle when t = 3 sec is
a.
105 m/sec
b.
27 m/sec
c.
108 m/sec
d.
42 m/s


Solution:

Here the displacement = S = ( 3t3 + 4t2 + 3t + 2) m.

Velocity of the particle = V = dS / dt
[Definition.]

= ddt( 3t3 + 4t2 + 3t + 2)
[Substitute S from step1.]

= 9t2 + 8t + 3
[Simplify.]

The velocity of the particle at t = 3 sec is 9(3)2 + 8(3) + 3 = 108 m/s.


Correct answer : (3)
 9.  
The displacement S of a particle in time t seconds is given by S = 7t3 - 3t2 - 4t + 3 ft . What is the acceleration of the particle at time t?
a.
(42t - 6) ft/s2
b.
(42t + 6) ft/s2
c.
(42t) ft/s2
d.
(- 42t - 6) ft/s2


Solution:

Here the displacement S = 7t3 - 3t2 - 4t + 3 ft.

The velocity of the particle at time t = V = dSdt
[Definition.]

= ddt ( 7t3 - 3t2 - 4t + 3)
[Substitute S from step1.]

Þ V = (21t2 - 6t - 4) ft/s.

The acceleration of the particle at time t = a = dVdt
[Definition.]

= ddt (21t2 - 6t - 4)
[Substitute V from step4.]

Þ a = (42t - 6) ft/s2
[Simplify.]


Correct answer : (1)
 10.  
What is the acceleration of a body at t = 4 sec whose displacement S(in metres) is related with the time as S = 3t3 - 5t2 - 4t + 2?
a.
- 62 m/sec2
b.
- 72m/sec2
c.
72 m/sec2
d.
62 m/sec2


Solution:

Here the displacement S = 3t3 - 5t2 - 4t + 2

The velocity of the particle at time t = V = dSdt
[Definition.]

= ddt( 3t3 - 5t2 - 4t + 2)
[Substitute S from step1.]

Þ V = 9t2 - 10t - 4

The acceleration of the body at time t = a = dVdt

= ddt (9t2 - 10t - 4)
[Substitute V from step4.]

Þ a = (18t - 10) m/sec2

At t = 4 sec, the acceleration = 18(4) - 10 = 72 - 10 = 62 m/sec2


Correct answer : (4)

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