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Motion - Acceleration

Grade 9CBSE

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

🔑Concepts

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Acceleration is defined as the rate of change of velocity of an object with respect to time. It is a vector quantity.

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If the velocity of an object changes from an initial value uu to a final value vv in time tt, the acceleration aa is given by a=v−uta = \frac{v - u}{t}.

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The SI unit of acceleration is m/s2m/s^2 or ms−2ms^{-2}.

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Uniform Acceleration: When an object travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time.

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Non-uniform Acceleration: When the velocity of an object changes by unequal amounts in equal intervals of time.

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Retardation (Deceleration): If the velocity of an object decreases with time, the acceleration is negative. This is called retardation or deceleration.

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On a velocity-time graph (v−tv-t graph), the slope of the line represents the acceleration of the object.

📐Formulae

a=v−uta = \frac{v - u}{t}

v=u+atv = u + at

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

v2−u2=2asv^2 - u^2 = 2as

Average Velocity (for uniform acceleration)=u+v2\text{Average Velocity (for uniform acceleration)} = \frac{u + v}{2}

💡Examples

Problem 1:

A car starts from rest and attains a velocity of 18 m/s18\ m/s in 6 s6\ s. Calculate its acceleration.

Solution:

Given: Initial velocity u=0 m/su = 0\ m/s, Final velocity v=18 m/sv = 18\ m/s, Time t=6 st = 6\ s. Using a=v−uta = \frac{v - u}{t}, we get a=18−06=3 m/s2a = \frac{18 - 0}{6} = 3\ m/s^2.

Explanation:

Since the car starts from rest, uu is zero. The change in velocity is divided by the time interval to find the rate of change.

Problem 2:

A bus moving at 20 m/s20\ m/s slows down to 5 m/s5\ m/s in 5 s5\ s after the brakes are applied. Find the acceleration.

Solution:

Given: u=20 m/su = 20\ m/s, v=5 m/sv = 5\ m/s, t=5 st = 5\ s. a=5−205=−155=−3 m/s2a = \frac{5 - 20}{5} = \frac{-15}{5} = -3\ m/s^2.

Explanation:

The negative sign indicates that the velocity is decreasing, which represents retardation or deceleration.

Problem 3:

A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h40\ km/h in 10 minutes10\ minutes. Find its acceleration in m/s2m/s^2.

Solution:

u=0 m/su = 0\ m/s. v=40 km/h=40×518=1009 m/sv = 40\ km/h = 40 \times \frac{5}{18} = \frac{100}{9}\ m/s. t=10 min=600 st = 10\ min = 600\ s. a=(100/9)−0600=1009×600=154≈0.0185 m/s2a = \frac{(100/9) - 0}{600} = \frac{100}{9 \times 600} = \frac{1}{54} \approx 0.0185\ m/s^2.

Explanation:

First, convert km/hkm/h to m/sm/s and minutes to seconds to maintain SI unit consistency before applying the formula.