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Graphical representation of motion equations Soumi Roy
24/03/2018 0 0
Equation of Motion (For Uniformly Accelerated Motion)

First Equation: v = u + at

Final velocity = Initial velocity + Acceleration × Time

Graphical Derivation

Suppose a body has initial velocity ‘u’ (i.e., velocity at time t = 0 sec.) at point ‘A’ and this velocity changes to ‘v’ at point ‘B’ in ‘t’ secs. i.e., final velocity will be ‘v’. For such a body there will be an acceleration.
a = Change in velocity/Change in Time
⇒ a = (OB - OA)/(OC-0) = (v-u)/(t-0)
⇒ a = (v-u)/t
⇒ v = u + at

Second Equation: s = ut + ½ at2

Distance travelled by object = Area of OABC (trapezium)
= Area of OADC (rectangle) + Area of â??ABD
= OA × AD + ½ × AD × BD
= u × t + ½ × t × (v – u)
= ut + ½ × t × at
⇒ s = ut + ½ at (â?µa = (v-u)/t)

Third Equation: v2 = u2 + 2as

s = Area of trapezium OABC ⇒ v2 = u2 + 2as
Example 1: A car starting from rest moves with uniform acceleration of 0.1 ms-2 for 4 mins. Find the speed and distance travelled.

Solution

u = 0 ms-1 (â?µ car is at rest)
a = 0.1 ms-2
t = 4 × 60 = 240 sec.
v = ?
From, v = u + at
v = 0 + (0.1 × 240)
⇒ v = 24 ms-1

Example 2: The brakes applied to a car produces deceleration of 6 ms -2 in opposite direction to the motion. If car requires 2 sec. to stop after application of brakes, calculate distance travelled by the car during this time.

Solution

Deceleration, a = − 6 ms-2
Time, t = 2 sec.
Distance, s = ?
Final velocity, v = 0 ms-1 (â?µ car comes to rest)
Now, v = u + at
Or  u = v – at
Or  u = 0 – (-6×2) = 12 ms-1
And, s = ut + ½at2
= 12 × 2 + ½ (-6 × 22)
= 24 – 12 = 12 m

Uniform Circular Motion → If a body is moving in a circular path with uniform speed, then it is said to be executing uniform circular motion.

→ In such a motion the speed may be same throughout the motion but its velocity (which is tangential) is different at each and every point of its motion. Thus, uniform circular motion is an accelerated motion.
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