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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 = $t$2 + 9$t$ + 64, then choose the acceleration of the particle from the following.
 a. $\frac{175}{4{s}^{3}}$ b. $\frac{175}{{s}^{2}}$ c. $\frac{175}{4{s}^{4}}$ d. $\frac{175}{{s}^{3}}$

#### 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 = 4$t$3 - 9$t$2 + 4$t$ + 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. $\frac{-11}{4}$ 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 = - 8$t$3 + 4$t$. 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 = 5$t$ - 2$t$3. 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 = 2$t$ - 2$t$2 ft . What is the velocity of the particle at any time $t$?
 a. (2 - 2$t$) ft/s b. 2 ft/s c. - 4 ft/s d. (2 - 4$t$) 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 = (4$t$3 - 7$t$2 + 4$t$ + 4) m. The instantaneous velocity of the body is
 a. (12$t$2 + 14$t$ + 4) m/s b. (12$t$2 - 14$t$ + 4) m/s c. (12$t$2 + 14$t$ - 4) m/s d. ( 12$t$2 + 14$t$) 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 = $a$$e$$\mathrm{6t}$ - $b$$e$-4$t$. Find the rate of change of displacement with respect to time $t$.
 a. 6$\mathrm{ae}$6$t$ + 4$\mathrm{be}$- 4$t$ b. $\mathrm{ae}$$\mathrm{6t}$ - $\mathrm{be}$-4$t$ c. $\mathrm{ae}$$\mathrm{6t}$ + $\mathrm{be}$-4$t$ d. 36$\mathrm{ae}$$\mathrm{6t}$ - 16$\mathrm{be}$-4$t$

#### 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 = 3$t$3 + 4$t$2 + 3$t$ + 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 = 7$t$3 - 3$t$2 - 4$t$ + 3 ft . What is the acceleration of the particle at time $t$?
 a. (42$t$ - 6) ft/s2 b. (42$t$ + 6) ft/s2 c. (42$t$) ft/s2 d. (- 42$t$ - 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 = 3$t$3 - 5$t$2 - 4$t$ + 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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