A common use of rate of change is to describe the motion of an object moving in a straight line. In such problems, it is customary to use either a horizontal or a vertical line with a designated origin to represent the line of motion.
On such lines,
The function s(t) that gives the position (relative to the origin) of an object as a function of time t is called a position function.
It is denoted by s = f (t) .
The velocity and the acceleration at time t is denoted as
v(t) = ds/dt
a(t) = d^{2}s/dt^{2}
Notes :
Point 1 :
Speed is the absolute value of velocity regardless of direction and hence
speed = |v(t)| = |ds/dt|
Point 2 :
Point 3 :
If t_{c} is the time point between the time points t_{1} and t_{2} (t_{1} < t_{c }< t_{1) }where the particle changes direction then the total distance travelled from time t_{1} to time t_{2} is calculated as
|s(t_{1}) - s(t_{c})| + |s(t_{c}) - s(t_{2})|
Problem 1 :
The temperature T in celsius in a long rod of length 10 m, insulated at both ends is the function of length x given by
T = x(10 - x)
Prove that the rate of change of temperature at the midpoint of the rod is 0.
Solution :
T = x(10 - x) or T(x) = x (10 - x)
The length of the rod is 10 m, to find the temperature at the midpoint of the rod is at x = 5.
To find the derivative, we use the product rule.
T'(x) = x(0 - 1) + (10 - x )(1)
= - x + 10 - x
T'(x) = -2x + 10
Temperature at midpoint :
T'(5) = -2(5) + 10
= -10 + 10
T'(5) = 0
So, temperature at the midpoint of the rod is 0.
Problem 2 :
A person learnt 100 words for an English test. The number of words the person remembers in t days after learning is given by
W(t) = 100×(1− 0.1t)^{2} , 0 ≤ t ≤10
What is the rate at which the person forgets the words 2 days after learning?
Solution :
Number of word the person forgets after 2 days,
W'(2) = -20(1 − 0.1(2))
= -20(1 - 0.2)
= -20(0.8)
= -16
So, the number of words he forgets is 16.
Problem 3 :
A particle moves so that the distance moved is according to the law
s(t) = t^{3}/3 - t^{2} + 3
At what time the velocity and acceleration is zero.
Solution :
s(t) = t^{3}/3 - t^{2} + 3
To find velocity, we will find the first derivative.
s'(t) = 3t^{2}/3 - 2t + 0
= t^{2} - 2t
When velocity = 0
t^{2} - 2t = 0
t(t - 2) = 0
t = 0 and t = 2
To find acceleration, we have to find the derivative of velocity function.
s'(t) = t^{2} - 2t
a(t) = s''(t) = 2t - 2
When acceleration = 0
2t - 2 = 0
t = 1
Problem 4 :
A particle is fired straight up from the ground to reach a height of s feet in t seconds where
s(t) = 128t - 16t^{2}
i) Compute the maximum height of the particle reached ?
ii) What is the velocity when the particle hits the ground ?
Solution :
s(t) = 128t - 16t^{2}
i) At the maximum height, the velocity v(t) of the particle is 0.
To find v(t), we have to find the derivative of s(t)
s'(t) = 128(1) - 16(2t)
v(t) = 128 - 32t
v(t) = 0
128 - 32t = 0
32t = 128
t = 128/32
t = 4
At 4 seconds, the particle reaches the maximum height.
Maximum height, when t = 4
s(t) = 128(4) - 16(4)^{2}
= 512 - 16(16)
= 512 - 256
= 256 ft
ii) When the particle hits the ground, when s = 0
0 = 128t - 16t^{2}
16t (8 - t) = 0
t = 0 and t = 8
It reaches the ground, when t = 0 and t = 8.
v(8) = 128 - 32(8)
= 128 - 256
= -128
The velocity when it reaches the ground is -128 ft/s
Problem 5 :
A particle moves along a horizontal line such that its position at any time t ≥ 0 is given by
s(t) = t^{3} − 6t^{2} + 9t + 1
where s is measured in meters and t in seconds?
(i) At what time the particle is at rest?
(ii) At what time the particle changes its direction?
(iii) Find the total distance travelled by the particle in the first 2 seconds.
Solution :
s(t) = t^{3} − 6t^{2} + 9t + 1
(i) When the particle is at rest, its velocity will become 0.
s(t) = t^{3} − 6t^{2} + 9t + 1
v(t) = 3t^{2} - 12t + 9 + 0
0 = 3t^{2} - 12t + 9
3(t^{2} - 4t + 3) = 0
3(t - 1)(t - 3) = 0
t = 1 and t = 3
(ii)
If 0 ≤ t < 1
v(t) = (t - 1)(t - 3)
say t = 0.5, then v(t) = + > 0
If 1 ≤ t < 3
say t = 2, then v(t) = - < 0
If t > 3
say t = 4, then v(t) = + > 0
Then the particle changes its direction in between 1 to 3 seconds.
(iii) The total distance travelled by the particle from time t = 0 to t = 2 is given by,
= |s(0) − s(1)| + |s(1) − s(2)|
=|1− 5 | + | 5 − 3|
= 6 meters
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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