Question

$\blue{\tt Question :}$
A car travelling at a velocity of 15 m/s^-1 due north speeds up uniformly to a velocity of 30 m/s^-1 seconds. Find the acceleration

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Answer 1

$\huge\red{\tt Correct \: Question :-}$

• A car travelling at a velocity of $\sf 15 \: ms^{-1}$ due north speeds up uniformly to a velocity of $\sf 30 \: ms^{-1}$ in 3 seconds. Find the acceleration.

$\huge\blue{\tt Given :-}$

• A car travelling at a velocity of $\sf 15 \: ms^{-1}$ due north speeds up uniformly to a velocity of $\sf 30 \: ms^{-1}$ in 3 seconds

$\huge\orange{\tt To \: Find :-}$

• The acceleration

$\huge\green{\tt Solution :-}$

We have,

• Final Velocity(v) = 30 m/s
• Initial Velocity(u) = 15 m/s
• Time taken(t) = 3 s

We know that,

$\boxed{\tt a = \dfrac{v - u}{t}}$

Where,

• a = Acceleration
• v = Final Velocity
• u = Initial Velocity
• t = Time taken

Substituting the given values we get,

$:\implies\tt a = \dfrac{30-15}{3}$

$:\implies\tt a = \dfrac{\cancel{15}}{\cancel3}$

$:\implies\orange{\tt a = 5 \: m/s^{2}}$

$\therefore\bf Acceleration \: is \: 5 \: m/s^{2}$

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Answer 2

Correct Question :

A car travelling at a velocity of 15 m/s due north speeds up uniformly to a velocity of 30 m/s in 3 seconds. Find the acceleration .

Given :

• Initial Velocity (u) = 15 m/s
• Final Velocity (v) = 30 m/s
• Time Taken (t) = 3 seconds.

To Find :

• Acceleration of the car .

Solution :

For Acceleration :

Using 1st Equation :

$\longmapsto\tt\boxed{v=u+at}$

Putting Values :

$\longmapsto\tt{30=15+a(3)}$

$\longmapsto\tt{30-15=3a}$

$\longmapsto\tt{15=3a}$

$\longmapsto\tt{a=\cancel\dfrac{15}{3}}$

$\longmapsto\tt\bf{a=5\:{m/s}^{2}}$

So , The Acceleration of the bus is 5 m/s ² ...

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Three Equations of Motion :

• v = u + at
• s = ut + 1/2 at²
• v² - u² = 2as

Here :

• u = Initial Velocity
• v = Final Velocity
• a = Acceleration
• s = Distance
• t = Time Taken

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