300963
Engineering
Physics
3.1
MOTION IN A STRAIGHT
LINE
Source: Young and Freedman, 2019
Motion in a straight line - Outline
In this section you will learn about displacement, velocity and acceleration in linear motion.
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Displacement
• Consider a particle at position x1 at time t1 and position x2 at time t2.
• The displacement (Δx) of the particle is defined as final position (x2) minus initial position (x1),
i.e.Δx = x2 – x1
• Units for displacement: metres (m)
Young and Freedman, 2019
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Example
A cheetah is crouched in ambush 20 m to the east of an observer. At time t = 0 s the cheetah
charges an antelope in a clearing 50 m east of the observer. The cheetah runs along a straight
line. During the first 2 s of the attack, the cheetah’s coordinate x varies with time according to
the equation x = 20 m + (5 m/s2)t2.
a) Find the displacement of the cheetah during the interval between t1 = 1 s and t2 = 2 s.
b) Find the average velocity during the same time interval.
c) Find the instantaneous velocity at t = 1 s.
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Diagram
Young and Freedman, 2019
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Solution
a) Displacement of the cheetah during the interval between t1 = 1 s and t2 = 2 s.
• At t1 = 1 s
𝑥1 = 20 + (5) (12) = 25 𝑚
• At t2 = 2 s
𝑥2 = 20 + (5) (22) = 40 𝑚
• Displacement;
Δ𝑥 = 𝑥2 – 𝑥1 = 40 – 25
= 15 𝑚
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Solution (continued)
b) Average velocity during the same time interval.
𝑣𝑎𝑣 = (𝑥2 – 𝑥1) / (𝑡2 – 𝑡1) = (40 – 25) / (2 − 1) = 15 𝑚/𝑠
c) Instantaneous velocity at t = 1 s.
𝑥 = 20 + (5)𝑡2
𝑣 = 𝑑𝑥/𝑑𝑡 = (5)(2𝑡) = 5 x 2 x 1 = 10 𝑚/𝑠
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Average and instantaneous acceleration
• Average acceleration
𝑎𝑎𝑣 = (𝑣2 – 𝑣1)/ (𝑡2 – 𝑡1) = Δ𝑣 / Δ𝑡
Units for average acceleration: m/s2
• Instantaneous acceleration
∆𝑣
𝑡→0 ∆𝑡
𝑎 = lim
=
𝑑𝑣
𝑑𝑡
Units for instantaneous acceleration: m/s2
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Let us do an example
Suppose the velocity v, of a car at any time is given by the equation;
𝑣 = 60 𝑚/𝑠 + (0.5 𝑚/𝑠3)𝑡2
a) Find the change in velocity of the car in the time interval between t1= 1 s and t2 = 3 s.
b) Find the average acceleration in this time interval.
c) Find the instantaneous acceleration at t = 3 s.
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Solution
a) Find the change in velocity of the car in the time interval between t1= 1 s and t2 = 3 s.
• At t1 = 1 s;
𝑣1 = 60 + 0.5 x 12
= 60.5 𝑚/𝑠
• At t2 = 3 s;
𝑣2 = 60 + 0.5 x 32
= 64.5 𝑚/𝑠
• Change in velocity;
Δ𝑣 = 𝑣2 – 𝑣1
= 64.5 – 60.5 = 4 𝑚/𝑠
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Solution (continued)
b) Average acceleration in this time interval.
𝑎𝑎𝑣 = (𝑣2 – 𝑣1 ) / (𝑡2 – 𝑡1)
= Δ𝑣 / Δ𝑡 = (64.5 – 60.5) / (3 – 1)
= 4/2 = 2 𝑚/𝑠2
c) Instantaneous acceleration at t = 3 s.
𝑣 = 60 + (0.5)𝑡2
𝑎 = 𝑑𝑣/𝑑𝑡 = (0.5)(2𝑡)
At t = 3 s;
𝑎 = (0.5 x 2 x 3) = 3 𝑚/𝑠2
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References
Young, H.D. and Freedman, R.A., 2019. University Physics with Modern Physics in SI Units. Pearson Education Canada.
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300963
Engineering
Physics