Question

A bus accelerates uniformly from 54 km/h to 72 km/h in 10 seconds Calculate
(i) acceleration in m/s2
(ii) distance covered by the bus in metres during this interval.

Answer 1

$\huge\bold{Solution}$

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Question:- A bus accelerates uniformly from 54 km/h to 72 km/h in 10 seconds Calculate

(i) acceleration in m/s2

(ii) distance covered by the bus in metres during this interval.

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First of all we should covert km/hr into m/s,

54km/hr => $\dfrac{54 × 1000m}{60 × 60s}$ => 15m/s

72km/hr => $\dfrac{72 × 1000m}{60 × 60s}$ => 20m/s

After that,

⭐Acceleration = $\dfrac{v - u}{\triangle{t}}$

a = $\dfrac{20 - 15}{10}$

a = $\bold{0.5ms^{-2}}$

⭐to find a distance travelled, using 2nd equation of motion...

S = ut + $\dfrac{1}{2}$at²

S = 15 × 10 + $\dfrac{1}{2}$0.5 × (10)²

S = 150 + 25

S = $\bold{175m}$

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HOPE IT HELPS YOU OUT♐♐

Answer 2

$<b><u><i>$

Heya !!

Here's ur solution -

Given,

u = 54km/h

v = 72km/h

t = 10sec

Step 1:

Convert km/h to m/s -

Calculating initial velocity (in m/s)

$54 \times 5 \div 18 \\ = 15ms { - }^{1}$

Calculating final velocity (in m/s)

$72 \times 5 \div 18 \\ = 20ms { - }^{1}$

Now,

We know that -

$accn \: = (v - u) \div t$

$=( 20 - 15) \div 10 \\ = 5 \div 10 \\ = 0.5ms { - }^{2}$

Hence, the acceleration is 0.5m/s²

Now,

We use 2nd equation of motion to find distance.

i.e. s = ut + 1/2at²

$s = 15 \times 10 + 0.5 \times 0.5 \times ( 10) {}^{2} \\ s = 150 + 0.25 \times 100 \\ s = 150 + 25 \\ s = 175m$

Distance covered is 175 metres.

Hope it helps :)