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Motion - Equations of Motion

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The equations of motion describe the behavior of a physical system in terms of its motion as a function of time.

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These equations are applicable only when an object moves in a straight line with uniform (constant) acceleration (aa).

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The variable uu denotes the initial velocity, vv denotes the final velocity, tt is the time interval, and ss is the distance or displacement covered.

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The first equation of motion, v=u+atv = u + at, represents the velocity-time relationship.

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The second equation of motion, s=ut+12at2s = ut + \frac{1}{2}at^2, represents the position-time relationship.

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The third equation of motion, v2−u2=2asv^2 - u^2 = 2as, represents the position-velocity relationship.

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If an object starts from rest, its initial velocity u=0u = 0. If an object comes to a stop (brakes applied), its final velocity v=0v = 0.

📐Formulae

v=u+atv = u + at

s=ut+12at2s = ut + \frac{1}{2}at^2

v2−u2=2asv^2 - u^2 = 2as

vavg=u+v2v_{avg} = \frac{u + v}{2}

💡Examples

Problem 1:

A racing car has a uniform acceleration of 4 m/s24\text{ m/s}^2. What distance will it cover in 10 s10\text{ s} after start?

Solution:

Given: u=0 m/su = 0\text{ m/s} (starts from rest), a=4 m/s2a = 4\text{ m/s}^2, t=10 st = 10\text{ s}. Using the second equation of motion: s=ut+12at2s = ut + \frac{1}{2}at^2. Substituting the values: s=(0×10)+12×4×(10)2=0+2×100=200 ms = (0 \times 10) + \frac{1}{2} \times 4 \times (10)^2 = 0 + 2 \times 100 = 200\text{ m}.

Explanation:

Since the car starts from rest, the initial velocity is zero. We apply the position-time relation to find the total distance covered during the period of constant acceleration.

Problem 2:

A stone is thrown in a vertically upward direction with a velocity of 5 m/s5\text{ m/s}. If the acceleration of the stone during its motion is 10 m/s210\text{ m/s}^2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Solution:

Given: u=5 m/su = 5\text{ m/s}, a=−10 m/s2a = -10\text{ m/s}^2 (downward), v=0v = 0 (at max height). Using v2−u2=2asv^2 - u^2 = 2as: 02−(5)2=2×(−10)×s⇒−25=−20s⇒s=1.25 m0^2 - (5)^2 = 2 \times (-10) \times s \Rightarrow -25 = -20s \Rightarrow s = 1.25\text{ m}. To find time, use v=u+atv = u + at: 0=5+(−10)t⇒10t=5⇒t=0.5 s0 = 5 + (-10)t \Rightarrow 10t = 5 \Rightarrow t = 0.5\text{ s}.

Explanation:

For upward motion, acceleration due to gravity is taken as negative because it opposes the direction of motion. At the highest point, the final velocity is always zero.