Third Equation of Motion


 
 
Concept Explanation
 

Third Equation of Motion

THIRD EQUATION OF MOTION:

large v^{2}-u^{2}=2as

v= Final Velocity

u = Initial Velocity

a= Acceleration

s= Distance

Let us derive the third equation of motion

Distance travelled=average velocity X time

large Average; Velocity = frac{v+ u}{2}

             large s=frac{u+v}{2}times t                            (I)

But from the first equation of motion, v = u + at

large at= v-u

large t= frac{v-u}{a}

Substituting the value of t in Eq (I)

  large therefore  large s=frac{u+v}{2}timesfrac{v-u}{a}

 or   large s=frac{v^{2}-u^{2}}{2a}

large v^{2}-u^{2}=2as

Example: A car is moving with an initial velocity of 5 m/s . The final velocity is 25 m/s when it accelerates at thr rate of 6 m/sq. s. Find the distance travelled

Sol: u = 5 m/s

       v= 25 m/s

       a = 6 m/sq. s

       s = ?

large v^{2}-u^{2}=2as

large 25^2-5^2= 2; X; 6; X; s

(25+5)(25-50 = 12 s

30 X 20 = 12 s

600 = 12 s

large s=frac{600}{12}= 50;m

The three equations listed can be used to solve the majority of kinematic problems.

Sample Questions
(More Questions for each concept available in Login)
Question : 1

A body is thrown vertically upwards and rises to a height of 10 m. the velocity with which the body was thrown upwards is (g=9.8m/s^2)

Right Option : C
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Explanation
Question : 2

The speed of a car reduces from  large 16 : m: s ^{-1}: : to : : 6: m: s^{-1} over a displacement of 8 m. What is the unifomr acceleration of the car ?

Right Option : A
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Explanation
Question : 3

A body with an initial velocity of 36 km/h accelerates uniformly at the rate of 18 cm s^{-2} over a distance of 200 m. calculate: the acceleration in ms^{-2} and its final velocity in ms^{-1}

Right Option : A
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Explanation
 
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