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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 = - t3 + 8t2 + 7t + 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.


Correct answer : (2)
 2.  
The distance travelled by a particle on a straight line in time t is given by s = 5 + 6t - t2 + 7t3 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.
50 7 ft/sec
d.
125 21 ft/sec
e.
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.


Correct answer : (4)
 3.  
The displacement s of a body with the time t is related as s = ae3t - be- 4t. What is the acceleration of the body at any time t?
a.
a - b
b.
ae3t + be- 4t
c.
9ae3t - 16be- 4t
d.
(3a - 4b)(e3t-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 .


Correct answer : (3)
 4.  
A particle moves along a straight line with the relation between the time and distance (s) in such a way that s = 9 - 5t2 + 10t3. At what time will the acceleration be 230 units/sec2?
a.
300 sec
b.
1 4 sec
c.
4 sec
d.
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.


Correct answer : (3)
 5.  
A particle covers the displacement s in time t which is given as s = k t12 where k is a constant, then find the acceleration of the particle at t = 4 sec.
a.
k16  units/sec2
b.
k units/sec2
c.
- k32  units/sec2
d.
k32  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.


Correct answer : (3)
 6.  
If the displacement s of a body is related with time t as s = 2 cos wt - 3 sin wt, then find the acceleration of the particle at time t.
a.
- s
b.
- ω2 s
c.
ω [- 2 cos ωt + 3 sin ωt]
d.
- ω2 [2 cos ωt + 3 sin ωt]
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.


Correct answer : (2)
 7.  
The displacement s of a moving particle is related with time as s = at2 + 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, 146 5and -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.


Correct answer : (2)
 8.  
The displacement of a body moving on straight line in time t is given as s = 256t - 16t33. 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.]


Correct answer : (3)
 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 = 112t - 16t2. 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.


Correct answer : (3)
 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.


Correct answer : (5)

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