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RECTILINEAR MOTION
Type of motion describes the movement of a particle or a body
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Rectilinear motion is simply the motion along a straight line
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This type of motion occurs in everyday life
whenever an object slides over a horizontal, low friction surface: e
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, a puck sliding across a
hockey rink
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𝒗𝒇 = 𝒗𝒊 + 𝒂𝒕
𝟏
𝒔 = 𝒗𝒊 𝒕 + 𝒂𝒕𝟐
𝟐
𝒗𝒇 ² = 𝒗𝒊 ² + 𝟐𝒂𝒔
𝒗𝟐 = 𝟐𝒈𝒉

FREE-FALLING BODY
All bodies in free-fall close to the Earth's surface accelerate vertically downwards with
the same acceleration: namely, g = 9
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𝒗 = 𝒈𝒕
𝒉=

𝟏 𝟐
𝒈𝒕
𝟐

Note: From constant acceleration, initial velocity = 0, final velocity = V, s = h and a = g
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𝒂=

𝒅𝒗
𝒅𝒕

𝒗=

𝒅𝒔
𝒅𝒕

𝒗𝒅𝒗 = 𝒂𝒅𝒔

Where:

Note:

𝑠 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑣 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑣𝑖 = 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑣𝑓 = 𝑓𝑖𝑛𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑎 = 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛
𝑔 = 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑑𝑢𝑒 𝑡𝑜 𝑔𝑟𝑎𝑣𝑖𝑡𝑦
𝑡 = 𝑡𝑖𝑚𝑒

• a is positive (+) if v is increasing (accelerate)
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• g is positive (+) if the particle is moving
downward
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PROBLEM SET
Problem 1
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38 m) high at the same instant that a
second ball is thrown upward from the ground with an initial velocity of 40 ft/sec (12
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When and where do they pass, and with what relative velocity?

Given:
ℎ= 80 ft
𝑣𝑓 = 0
𝑣𝑖 = 40 ft/s
𝑔= 32
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1𝑡 2 + (40𝑡 − 16
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2)𝑡 2
ℎ1 = 16
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4 𝑓𝑡

2

From B to C (upward motion)

They pass each other after 2 seconds at 64
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Answer

𝟏

From the formula 𝒔 = 𝒗𝒊 𝒕 + 𝟐 𝒂𝒕𝟐

Velocity at C of stone from A (after 2 seconds)

𝑣𝑐1 = gt = 32
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4 ft per sec

1

ℎ2 = 𝑣𝑖 𝑡 − 2 𝑔𝑡 2
1

ℎ2 = 40𝑡 − 2 (32
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1𝑡

Velocity at C of stone from B (after 2 seconds)

2

𝑣𝑐2 = 𝑣𝑖 − 𝑔𝑡 = 40 − 32
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4
Relative velocity

𝑣𝑟 = 𝑣𝑐1 + 𝑣𝑐2
𝑣𝑟 = 64
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4
𝑣𝑟 = 40 ft per sec answer

𝑓𝑡
𝑠𝑒𝑐

(Downward)

Problem 2
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Determine the speed at which it hits the ground and the time travelled
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2 𝑓𝑡 𝑝𝑒𝑟 𝑠𝑒𝑐²

𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡

𝑣𝑓 ² = 𝑣𝑖 ² + 2𝑎𝑠

59
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2)𝑡

𝑣𝑓 ² = (18)2 + 2(32
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532 𝑓𝑡/𝑠𝑒𝑐

answer
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3 𝑠𝑒𝑐 answer
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Traveling with an initial speed of 70 km/h, a car accelerates at 6000 km/h² along a
straight road
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792 𝑘𝑚 = 792 𝑚 answer
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Problem 5
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At what upward velocity must Iverson leave the floor
to achieve this?
Given:

Solution:

𝑠 = 1𝑚

𝑣𝑓 ² = 𝑣𝑖 ² + 2𝑎𝑠

𝑎 = 𝑔 = −9
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81)(1)

Find 𝑣𝑖

𝑣𝑖 = 4
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Problem 6
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32 m) during the 10th sec of its
motion and 18 ft (5
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Find its initial velocity and its
constant acceleration

Solution:
𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡
From (1):

24 = 𝑣𝑖 + 10𝑎

(1)

18 = 𝑣𝑖 + 12𝑎

(2)

24 = 𝑣𝑖 + 10(−3)
𝑣𝑖 = 54 𝑓𝑡/𝑠𝑒𝑐

Subtract (1) to (2)
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answer
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An airplane lands with initial velocity of 70 m/s and then decelerates at 1
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What is its final velocity?

Given:
𝑣𝑖 = 70 𝑚/𝑠
𝑎 = −1
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50)(40) = 10 𝑚/𝑠𝑒𝑐

answer
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Dragsters can achieve an average acceleration of 26
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Suppose a dragster
accelerates from the rest at this rate for 5
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How far does it travel in this time?
𝑣𝑖

𝑠𝑖

𝑣𝑓

𝑎 = 26
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56 sec

𝑠=

1 2
𝑎𝑡
2

𝑠=

1
(26)(5
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Problem 9
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If its initial velocity is
10 m/s and it accelerates at 2
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𝑣𝑓 =?

Problem 10
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a
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0 s?
b
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8 𝑚/𝑠²

Solution:
(a)
𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡
𝑣𝑓 = 𝑣𝑖 + (−9
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(b)
1
𝑠 = 𝑣𝑖 𝑡 + 𝑎𝑡 2
2
1

𝑠 = 0 + 2 (−9
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the bricks falls 78 meters
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Problem 11
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When at a height of 5
km and velocity of 200 m/s, it releases its booster
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Solution:

(a)
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8 𝑚/𝑠²
𝑣 2 = 𝑣𝑖2 + 2𝑔ℎ
𝑣²

(200)²

ℎ = 2𝑔𝑖 = 2(−9
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8 𝑚

answer
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𝑣 2 = 𝑣𝑖2 + 2𝑔ℎ
𝑚

𝑣 2 = (200𝑚)2 + (−9
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8 𝑚/𝑠

answer
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A stone is dropped from a captive balloon at an elevation of 1000ft
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if g =
32ft/s²
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1
𝑠 = 𝑣𝑖2 𝑡2 − 𝑔𝑡2 2
2
1

ℎ2 = 248𝑡2 − 2 (32)𝑡2 ²
ℎ2 = 248(𝑡1 − 2) − 16(𝑡1 − 2)²

ℎ = ℎ1 + ℎ2
1000 = 16𝑡12 + [248(𝑡1 − 2) − 16(𝑡1 − 2)2 ]
1560 = 312𝑡1
𝑡1 = 5𝑠𝑒𝑐

answer
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Problem 13
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With what speed in meters per
second does it strike the ground?
Solution:

vf2 – v02 = + 2gh
vf2 – 0 = 2(9
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1
5m

vf = 9
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Problem 14
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12–2 moves in a straight line such that for a
short time its velocity is defined by v = (3t2 + 2t) ft/s, where t is in seconds
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When t = 0, s = 0
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Since v = f (t), the acceleration
is determined from a = dv/dt, since this
equation relates a, v, and t
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Problem 15
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If the mass of the ball is 10 gm after what
time will the ball strike the earth?

100m

Solution:
NOTE: since the ball is dropped the initial velocity is zero
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81 𝑠2 ) 𝑡²
𝑡 = 4
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EXERCISES:
1
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The
police car immediately gives chase, accelerating uniformly to reach a speed
of 90 km/h in10 s and continues at this speed until he overtakes the other car
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2
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One of them, colored red, accelerates at
0
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The other car, colored blue,
accelerates at 0
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Draw both journeys on the same velocity-time graph and determine the time and distance that
the second car overtakes the first car

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