Derive the relation s = u t + 1 2 a t 2 , where the symbols have their usual meanings.

Using the formulae of acceleration We know that, a = d v / d t ∫ v u d v = ∫ a 0 d t ⇒ v - u = a t ⇒ v = u + a t . . . . . . ( 1 ) a = v d v / d s ∫ u v v d v = ∫ o s a d s ⇒ v 2 - u 2 = 2 a s . . . . . ( 2 ) P u t t i n g v = u + a t i n e q u a t i o n v 2 - u 2 = 2 a s u + a t 2 - u 2 = 2 a s W e g e t s = u t + a t 2 2 .

Discuss velocity - time graph. Derive relation s = 4 t + 1 2 at 2 from this graph.

s = 4 t + 1 2 at 2 ⇒ v = d s d t = d 4 t + 1 2 a t 2 d t = 4 + 1 2 × 2 × a t ⇒ v = 4 + a t ⇒ v = a t + 4 Since, acceleration is constant. It will follow the the trajectory of y = m x + c where c is the intercept on Y-axis.

E M B I B E

Bureau's Higher Secondary Physics Class 11>Chapter 3 - Kinematics>MODEL QUESTIONS>Q 1

EASY

11th Odisha Board

IMPORTANT

Earn 100

Using calculus method derive the expression for the displacement in the nth second of a body moving with uniform acceleration along a straight line.

Important Questions on Kinematics

EASY

11th Odisha Board

IMPORTANT

Bureau's Higher Secondary Physics Class 11>Chapter 3 - Kinematics>MODEL QUESTIONS>Q 2

Derive the relation

s = u t + \frac{1}{2} a t^{2}

, where the symbols have their usual meanings.

EASY

11th Odisha Board

IMPORTANT

Bureau's Higher Secondary Physics Class 11>Chapter 3 - Kinematics>MODEL QUESTIONS>Q 3

Obtain velocity-time and position-time relation for the uniform and uniformly accelerated motion. Explain the results with the help of position - time and velocity - time graph for both cases.

HARD

11th Odisha Board

IMPORTANT

Bureau's Higher Secondary Physics Class 11>Chapter 3 - Kinematics>MODEL QUESTIONS>Q 4

Using calculus method prove that