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Understanding Motion through Experience - Equations of Motion-advanced

Grade 9CBSE

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

🔑Concepts

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Equations of motion are valid only for objects moving with a constant (uniform) acceleration aa along a straight line.

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The first equation v=u+atv = u + at defines the velocity-time relationship, where uu is initial velocity and vv is final velocity.

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The second equation s=ut+12at2s = ut + \frac{1}{2}at^2 defines the position-time relationship, where ss is the displacement.

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The third equation v2−u2=2asv^2 - u^2 = 2as defines the position-velocity relationship, useful when time tt is not given.

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The displacement in a specific nthn^{th} second is calculated using Sn=u+a2(2n−1)S_{n} = u + \frac{a}{2}(2n - 1).

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For objects in free fall, acceleration aa is replaced by gg (acceleration due to gravity), approximately 9.8 m/s29.8 \text{ m/s}^2 or 10 m/s210 \text{ m/s}^2. Upward motion takes −g-g and downward motion takes +g+g.

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The area under a Velocity-Time graph represents the displacement ss, while the slope represents acceleration aa.

📐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}

Sn=u+a2(2n−1)S_n = u + \frac{a}{2}(2n - 1)

💡Examples

Problem 1:

A car traveling at 20 m/s20 \text{ m/s} applies brakes and comes to a stop in 5 seconds5 \text{ seconds}. Calculate the acceleration and the distance traveled during this time.

Solution:

Given: Initial velocity u=20 m/su = 20 \text{ m/s}, final velocity v=0 m/sv = 0 \text{ m/s}, and time t=5 st = 5 \text{ s}. Step 1: Find acceleration using v=u+atv = u + at 0=20+a(5)0 = 20 + a(5) −20=5a-20 = 5a a=−4 m/s2a = -4 \text{ m/s}^2 Step 2: Find distance using s=ut+12at2s = ut + \frac{1}{2}at^2 s=(20)(5)+12(−4)(5)2s = (20)(5) + \frac{1}{2}(-4)(5)^2 s=100−2(25)s = 100 - 2(25) s=100−50=50 ms = 100 - 50 = 50 \text{ m}.

Explanation:

The negative sign in acceleration indicates deceleration (retardation). The second equation helps find the total displacement during the braking period.

Problem 2:

A ball is thrown vertically upwards with a velocity of 25 m/s25 \text{ m/s}. Calculate the maximum height reached by the ball. (Take g=10 m/s2g = 10 \text{ m/s}^2)

Solution:

Given: u=25 m/su = 25 \text{ m/s}, v=0v = 0 (at peak), a=−g=−10 m/s2a = -g = -10 \text{ m/s}^2. Using the third equation: v2−u2=2asv^2 - u^2 = 2as 02−(25)2=2(−10)s0^2 - (25)^2 = 2(-10)s −625=−20s-625 = -20s s=62520=31.25 ms = \frac{625}{20} = 31.25 \text{ m}.

Explanation:

At the highest point, the final velocity is always zero. Since the ball moves against gravity, the acceleration is taken as negative.

Problem 3:

An object starts from rest and moves with a uniform acceleration of 2 m/s22 \text{ m/s}^2. Find the distance traveled by the object in the 10th10^{th} second.

Solution:

Given: u=0u = 0, a=2 m/s2a = 2 \text{ m/s}^2, n=10n = 10. Using the formula for distance in nthn^{th} second: Sn=u+a2(2n−1)S_n = u + \frac{a}{2}(2n - 1) S10=0+22(2×10−1)S_{10} = 0 + \frac{2}{2}(2 \times 10 - 1) S10=1×(20−1)S_{10} = 1 \times (20 - 1) S10=19 mS_{10} = 19 \text{ m}.

Explanation:

This specific formula is used to find the displacement during one particular second (the 10th), rather than the total displacement over 10 seconds.

Problem 4:

A body accelerates from 15 m/s15 \text{ m/s} to 25 m/s25 \text{ m/s} covering a certain distance. If the acceleration is 2 m/s22 \text{ m/s}^2, find the distance using vertical subtraction for the velocity squares.

Solution:

Given: u=15u = 15, v=25v = 25, a=2a = 2. Formula: v2−u2=2asv^2 - u^2 = 2as v2=252=625v^2 = 25^2 = 625 u2=152=225u^2 = 15^2 = 225 To find v2−u2v^2 - u^2: 625−225400\begin{array}{r} 625 \\ -225 \\ \hline 400 \end{array} Now, 400=2(2)s400 = 2(2)s 400=4s400 = 4s s=100 ms = 100 \text{ m}.

Explanation:

The difference between the squares of final and initial velocity is proportional to the product of acceleration and displacement.