Question

A car acquires a velocity of 72 kmh-1in 10s starting from rest. Find a.The acceleration b.The average velocity

Answer 1

Answer:

velocity =20 acceleration =2

Explanation:

1kmh=5/18m/s

hence

72kmh=72×5/18=20m/s

avg. velocity =20m/s

acceleration=velocity ÷time

=20÷10=2m/s^2

Answer 2

Answer

a . a = 2 m/s²

b . Average velocity = 10 m/s

Given

A car acquires a velocity of 72 km/h in 10 s starting from rest

To Find

a . acceleration

b . Average velocity

Concept Used

$\boxed{\begin{minipage}{5cm} \underbrace{\bf Equations\ of\ motion}\ :\\\\\rm \to\ v=u+at\\\\\rm \to\ s=ut+\dfrac{1}{2}at^2\\\\\rm \to\ v^2-u^2=2as\\\\\rm \to\ S_n=u+\dfrac{a}{2}(2n-1) \end{minipage}}$

SI unit of velocity is m/s

SI unit acceleration is m/s²

SI unit of time is second

SI unit of displacement is meter

Average Velocity :

It is defined as net displacement per total time

Solution

Initial velocity , u = 0 m/s

[ ∵ starting from rest ]

Final velocity , v = 72 km/h

So , we need to convert from km/h to m/s .

$\to \rm v=72 \times \dfrac{5}{18}\\\\\to \rm v=4\times 5\\\\\to \bf v=20\ m/s$

Time , t = 10s

Apply 1st equation of motion ,

⇒ v = u + at

⇒ (20) = (0) + a(10)

⇒ 20 = 0 + 10a

⇒ 10a = 20

⇒ a = 2 m/s²

So , acceleration , a = 2m/s²

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Apply 3rd equation of motion ,

⇒ v² - u² = 2as

⇒ (20)² - (0)² = 2(2)s

⇒ 400 - 0 = 4s

⇒ 400 = 4s

⇒ 4s = 400

⇒ s = 100 m

So , net displacement , s = 100 m

Time , t = 10 s

Apply formula for average velocity ,

$\to \rm A_{velocity}=\dfrac{Net\ displacement}{Time}\\\\\to \rm A_{velocity}=\dfrac{100}{10}\\\\\to \rm A_{velocity}=10\ m/s$

So , Average velocity = 10 m/s

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