# Rate of Change Worksheet

Rate of Change Worksheet
• Page 1
1.
The displacement (in meters) of a particle in time $t$ is given by $s$ = - $t$3 + 8$t$2 + 7$t$ + 2. Find the velocity of the particle when $t$ = 2 sec.
 a. 40 m/s b. 27 m/s c. 80 m/s d. 28 m/s e. 76 m/s

#### Solution:

Displacement of the particle, s = - t3 + 8t2 + 7t + 2
[Write the function.]

v = dsdt = ddt (- t3 + 8t2 + 7t + 2)
[Find the velocity of the particle, v = dsdt.]

= - 3t2 + 16t + 7
[Differentiate with respect to t.]

The velocity of the particle at t = 2 sec is - 3(2)2 + 16(2) + 7 = 27
[Substitute t = 2 and simplify.]

Therefore, the velocity of the particle at t = 2 sec is 27 m/s.

2.
The distance travelled by a particle on a straight line in time $t$ is given by $s$ = 5 + 6$t$ - $t$2 + 7$t$3 where $s$ is in feet and $t$ is in seconds. Find the velocity of the particle when the acceleration is zero.
 a. 5 ft/sec b. 6 ft/sec c. $\frac{50}{7}$ ft/sec d. $\frac{125}{21}$ ft/sec e. $\frac{43}{7}$ ft/sec

#### Solution:

Displacement of the particle, s = 5 + 6t - t2 + 7t3
[Write the function.]

v = dsdt = ddt (5 + 6t - t2 + 7t3)
[Find the velocity of the particle, v = dsdt.]

= 6 - 2t + 21t2
[Differentiate with respect to t.]

a = dvdt = ddt (6 - 2t + 21t2)
[Find the acceleration of the particle, a = dvdt.]

= - 2 + 42t
[Differentiate with respect to t.]

If the acceleration of the particle, a = 0, then - 2 + 42t = 0

t = 1 / 21
[Solve for t.]

So, the acceleration of the particle is zero at t = 1 / 21 sec.

At t = 1 / 21 sec, the velocity of the particle = 6 - 2(1 / 21) + 21(1 / 21)2
[Substitute t = 1 / 21 in 6 - 2t + 21t2.]

Therefore, the velocity of the particle when acceleration is zero is 125 / 21 ft/sec.

3.
The displacement $s$ of a body with the time $t$ is related as $s$ = $\mathrm{ae}$3$t$ - $\mathrm{be}$- 4$t$. What is the acceleration of the body at any time $t$?
 a. $a$ - $b$ b. $\mathrm{ae}$3$t$ + $\mathrm{be}$- 4$t$ c. 9$\mathrm{ae}$3$t$ - 16$\mathrm{be}$- 4$t$ d. (3$a$ - 4$b$)(${e}^{3t}-{e}^{-4t}$)

#### Solution:

Displacement of the body, s = ae3t - be- 4t
[Write the function.]

v = dsdt = ddt (ae3t - be- 4t)
[Find the velocity of the body, v = dsdt.]

= 3ae3t - 4be- 4t(- 1)
[Differentiate with respect to t.]

= 3ae3t + 4be- 4t

a = dvdt = ddt (3ae3t + 4be- 4t)
[Find the acceleration of the body, a = dvdt.]

= 9ae3t + 16be- 4t(- 1)
[Differentiate with respect to t.]

= 9ae3t - 16be- 4t

Therefore, the acceleration of the body at any time t is 9ae3t - 16be- 4t .

4.
A particle moves along a straight line with the relation between the time and distance ($s$) in such a way that $s$ = 9 - 5$t$2 + 10$t$3. At what time will the acceleration be 230 units/sec2?
 a. 300 sec b. $\frac{1}{4}$ sec c. 4 sec d. $\frac{11}{3}$ sec e. 280 sec

#### Solution:

s = 9 - 5t2 + 10t3

v = dsdt = ddt(9 - 5t2 + 10t3)
[Find the velocity of the particle, v = dsdt.]

= - 10t + 30t2
[Use Difference, Sum Rule.]

a = dvdt = ddt(- 10t + 30t2)
[Find the acceleration of the particle, a = dvdt.]

= - 10 + 60t
[Use Difference, Sum Rule.]

If the acceleration of the particle is 230 units/sec2, then - 10 + 60t = 230

t = 4
[Solve for t.]

Therefore, the acceleration of the particle is 230 units/sec2 at time t = 4 sec.

5.
A particle covers the displacement $s$ in time $t$ which is given as $s$ = $k$ ${t}^{\frac{1}{2}}$ where $k$ is a constant, then find the acceleration of the particle at $t$ = 4 sec.
 a. units/sec2 b. $k$ units/sec2 c. - units/sec2 d. units/sec2

#### Solution:

The displacement of the particle, s = k t12
[Write the function.]

v = dsdt = ddt (k t12)
[Find the velocity of the particle, v = dsdt.]

= k (1 / 2 t-12)
[Differentiate with respect to t.]

a = dvdt = ddt (k2 t-12)
[Find the acceleration of the particle, a = dvdt.]

= k2 (- 1 / 2 t-32)
[Differentiate with respect to t.]

The acceleration of the particle at t = 4 sec is k2 [- 1 / 2 (4)-32]

= - k32

Therefore, the acceleration of the particle at t = 4 sec is - k32  units/sec2.

6.
If the displacement $s$ of a body is related with time $t$ as $s$ = 2 cos w$t$ - 3 sin w$t$, then find the acceleration of the particle at time $t$.
 a. - $s$ b. - ω2 $s$ c. ω [- 2 cos ω$t$ + 3 sin ω$t$] d. - ω2 [2 cos ω$t$ + 3 sin ω$t$] e. ω2 $s$

#### Solution:

The displacement of the body, s = 2 cos wt - 3 sin wt
[Write the function.]

v = dsdt = ddt (2 cos wt - 3 sin wt)
[Find the velocity of the body, v = dsdt.]

= - 2 (sin wt)(w) - 3 (cos wt)(w)
[Differentiate with respect to t.]

= - w[2 sin wt + 3 cos wt]
[Factor.]

a = dvdt = ddt[- w(2 sin wt + 3 cos wt)]
[Find the acceleration of the body, a = dvdt.]

= - w[2 (cos wt)(w) - 3 (sin wt)(w)]
[Differentiate with respect to t.]

= - w2[2 cos wt - 3 sin wt]
[Factor.]

= - w2 s
[Replace 2 cos wt - 3 sin wt with s.]

Therfore, the acceleration of the body at time t is - w2 s.

7.
The displacement $s$ of a moving particle is related with time as $s$ = $\mathrm{at}$2 + $\mathrm{bt}$ + $c$. If the displacement after 1 sec is 20 m, the velocity after 2 sec is 30 m/sec and acceleration is 10 m/sec2, then find the values of $a$, $b$ and $c$.
 a. 5, $\frac{146}{5}$and $\frac{-47}{5}$ b. 5, 10 and 5 c. 5, -10 and -5 d. 20, -50 and -10 e. 5, 50 and 5

#### Solution:

The displacement of the particle, s = at2 + bt + c
[Write the function.]

At t = 1 sec, the displacement is 20 m.

a(1)2 + b(1) + c = 20, 1a + 1b + c = 20 ------ (1)
[Substitute t = 1.]

v = dsdt = ddt(at2 + bt + c)
[Find the velocity of the particle, v = dsdt.]

= 2at + b
[Differentiate with respect to t.]

The velocity of the particle after 2 sec is 30 m/sec.

2a(2) + b = 30, 4a + b = 30 ------ (2)
[Substitute t = 2 sec.]

Acceleration of the particle, dvdt = ddt (2at + b) = 2a

The acceleration of the particle after 2 sec is 10 m/sec2.

2a = 10, a = 5
[Solve for a.]

4(5) + b = 30 b = 10
[Substitute a = 5 in equation (2).]

1(5) + 1(10) + c = 20 c = 5
[Substitute a = 5, b = 10 in equation (1).]

Therefore, the values of a, b and c are 5, 10 and 5 respectively.

8.
The displacement of a body moving on straight line in time $t$ is given as $s$ = 256$t$ - $\frac{16{t}^{3}}{3}$. At what instant will the velocity of the body vanishes?
 a. 256 sec b. 384 sec c. 4 sec d. 16 sec e. 19.59 sec

#### Solution:

The displacement of the body, s = 256t - 16t33
[Write the function.]

v = dsdt = ddt(256t - 16t33)
[Find the velocity of the body, v = dsdt.]

= 256 - 16 / 3 (3t2) = 256 - 16t2
[Differentiate with respect to t.]

If the velocity of the body vanishes, then v = 0.

256 - 16t2 = 0

t = - 4, 4
[Solve for t.]

At t = 4 sec, the velocity vanishes.
[Time is non-negative.]

9.
A projectile is fired straight up from ground level with an initial velocity of 112 feet per second. Its height $s$ above the ground after time $t$ is given by $s$ = 112$t$ - 16$t$2. What is its velocity and acceleration at $t$ = 2 sec?
 a. - 32 ft/sec, 48 ft/sec2 b. 96 ft/sec, - 16 ft/sec2 c. 48 ft/sec, - 32ft/sec2 d. 16 ft/sec, 32 ft/sec2 e. 80 ft/sec, - 16 ft/sec2

#### Solution:

s = 112t - 16t2
[Write the function.]

v = dsdt = ddt(112t - 16t2)
[Find the velocity of the stone, v = dsdt.]

= 112 - 32t
[Differentiate with respect to t.]

The velocity of the stone at t = 2 sec is 112 - 32(2) = 48
[Substitute t = 2 sec.]

a = dvdt = ddt(112 - 32t)
[Find the acceleration of the stone, a = dvdt.]

= - 32
[Differentiate with respect to t.]

The acceleration at t = 2 sec is - 32 ft/sec2.

Therefore, after 2 seconds, the velocity, acceleration of the stone projected vertically upwards with an initial velocity of 112 ft/sec are 48 ft/sec and - 32 ft/sec2.

10.
The displacement (in feet) of the particle in time $t$ is given by $s$ = ln ($t$ + 3). What is the velocity of the particle at time $t$ = 2.6 sec?
 a. 1.722 ft/sec b. 1.386 c. 0.535 ft/sec d. 0.748 ft/sec e. 0.178 ft/sec

#### Solution:

The displacement of the particle, s = ln (t + 3)
[Write the function.]

v = dsdt = ddt(ln (t + 3))
[Find the velocity of the particle, v = dsdt.]

= 1t+3
[Differentiate with respect to t.]

The velocity of the particle at t = 2.6 sec is 12.6+3
[Substitute t = 2.6.]

0.178 ft/sec
[Simplify.]

The velocity of the particle at t = 2.6 sec is approximately 0.178 ft/sec.