Question

A car accelerates uniformly from 18km/h to 36km/h in 5 sec i)calculate the acceleration and ii)the distance covered by car in that time
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Answer 1

• Initial velocity of the car, u = 18 km/hr

$1 \: \dfrac{km}{hr} \: = \dfrac{5}{18} \: \dfrac{m}{s} \\ \\ 18 \: \dfrac{km}{hr} \: = 18 \times \dfrac{5}{18} \: \dfrac{m}{s} \\ \\u = 5 \: \dfrac{m}{s}$

• Final velocity of the car, v = 36 km/hr

$1\: \dfrac{km}{hr} \: = \dfrac{5}{18} \: \dfrac{m}{s} \\ \\36\: \dfrac{km}{hr} \: = 36 \times \dfrac{5}{18} \: \dfrac{m}{s} \\ \\v = 2 \times 5 \: \dfrac{m}{s} \\ \\ v = 10 \: \dfrac{m}{s}$

• Time, t = 5 sec

• Let a be the acceleration

• Let s be the distance covered.

According to the first equation of motion.,

v = u + at

$10 = 5 +( a \times 5) \\ \\ 10 - 5 = 5a \\ \\ 5 = 5a \\ \\ a = \dfrac{5}{5} \\ \\ a = 1 \: \dfrac{m}{ {s}^{2} }$

The acceleration of the car is 1 m/s².

According to third equation of motion,

${v}^{2} - {u}^{2} = 2as \\ \\ {(10)}^{2} - {(5)}^{2} = 2 \times 1 \times s \\ \\ 100 - 25 = 2s \\ \\ 75 = 2s \\ \\ s = \frac{75}{2} \\ \\ s = 37.5 \: m$

Distance travelled by the car is 37.5 m.

Answer 2

Answer:

Given :-

• Initial velocity = 18 × 5/18 = 5 m/s
• Final velocity = 36 × 5/18 = 10 m/s
• Time taken = 5 second

To Find :-

1. Acceleration
2. Distance covered

Formulas to be applied :-

For acceleration

$\sf \: v = u + at$

For Distance travelled

$\sf \: {v}^{2} - {u}^{2} = 2(a)(s)$

Solution :-

Let's find acceleration

$\tt \implies \: 10 = 5 + a(5)$

$\tt \implies \: 10 = 5 + 5a$

$\tt \implies \: 10 - 5 = 5a$

$\tt \implies \: 5 = 5a$

$\frak \red {\implies \: a = 1 \: ms {}^{ - 1} }$

Let's find Distance travelled

$\tt \implies {10}^{2} - {5}^{2} = 2(1)(s)$

$\tt \implies 100 - 25 = 2s$

$\tt \implies \: 75 = 2s$

$\tt \implies \: s = \cancel\dfrac{75}{2}$

$\frak \blue{s = 37.5 \: m}$

Hence :-

Acceleration is 1 m/s and Distance travelled is 37.5 m.