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Dynamic Engineering System - Linear Motion Notes
Transcript of Dynamic Engineering System - Linear Motion Notes
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Linear MotionMotion with constant acceleration
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Velocity
Time
u
u = initial velocity
v
v = final velocityGradient
gradient = acceleration
acceleration = a
v u
Rearrange equation:
t
1
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Velocity
Time
u
u = initial velocity
v
v = final velocity
v u
Rearrange equation:
t
area
under = distance
graph
distance = s
Area 1
area
under = area 1 + area 2
graph
Area 2
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We have:
We also have:
So,
Rearrange equation:
2
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We have:
We also have:
So,
Rearrange equation:
3
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3 Main Equations:
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Alternative method
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Alternative method
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Example 1 - Question
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Example 2 -Question
} Find the deceleration of a car that isbrought to rest in 60 m from a speed of 45
km/h. What is the time taken?
} t = 9.6 s #
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Example 3 - Question
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Example 4 - Question
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Example 4 - Solution
} s = 133.3 m. The car hit the cow becausewith the given retardation, the car will
stop after 133.3 m #
} v = 16 m/s. The car hit the cow becauseafter 50 m, the car still have velocity, thatmeans the car is still moving #
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Example 5 - Question
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Revision
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Revision
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Linear MotionVertical motion under gravity
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Acceleration due gravity = g
If an object moves
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Example 1 - Question
}An object is projected vertically upwardswith an initial velocity of 40 m/s. Calculate
(a) the maximum height reached, (b) thetime taken to reach the maximum height,(c) the time taken for the object to fallback to its initial point of projection.
} s = 81.5 m #
} t = 4.1 s #
} t = 4.1 s #
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Revision
}A brick falls off the top of building. If theheight of the building is 20 m, determine
the time the bricks takes to reach theground.
}A ball is thrown vertically upwards with avelocity of 15 m/s. Determine the greatest
height reached above the point ofprojection.
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Revision
}A ball is thrown vertically downwards withan initial velocity with an initial velocity of
4 m/s from the edge of a cliff. If the ballhits the ground at the base of the cliffafter 2s, determine the height of the cliff.