Question

A car  accelerates uniformly from 36km/h to 54 km/h in 10 seconds. Find the acceleration and the distance travelled by the  car .​

Answer 1

Explanation:

$\bf\underline{\underline{\red{Given:-}}}$

• Initial velocity = 36km/h

• Final velocity = 54km/h

• Time = 10sec

$\bf\underline{\underline{\blue{To\:Find:-}}}$

• The acceleration.

• The distance travelled by the car.

$\bf\underline{\underline{\green{Solution:-}}}$

Converting km/h to m/s

To convert km/h to m/s we have to multiply it with 5/18.

$\sf Initial \: velocity = 36km/h$

$\sf = 36 \times \dfrac{5}{18}$

$\sf = 2 \times 5$

$\sf = 10m/s$

$\sf Initial \: velocity = 10m/s$

$\sf Finall \: velocity = 54km/h$

$\sf = 54 \times \dfrac{5}{18}$

$\sf = 3 \times 5$

$\sf = 15m/s$

$\sf Final \: velocity = 15m/s$

$\sf We \: have :-$

• $\sf Initial\: velocity (u) = 10m/s$

• $\sf Final\: velocity (v) = 15m/s$

• $\sf Time (t) = 10\: sec$

$\sf Now, \: Let\: us \:find \: the\: acceleration$

$\sf According\: to\: the\: first \: equation\: of \: motion:-$

$\boxed{\sf \leadsto v = u + at}$

$\sf \implies 15 = 10 + a(10)$

$\sf \implies 15 - 10 = 10a$

$\sf \implies 5 = 10a$

$\sf \implies a = \dfrac{5}{10} = 0.5$

$\underline{\boxed{\pink{ \sf \therefore Acceleration = 0.5m/s^{2}}}}$

Now:-

$\sf According\: to\: the\: third\: equation\: of \: motion:-$

$\boxed{\sf \leadsto s = ut + \dfrac{1}{2}a{t}^{2}}$

$\sf\implies s = 10 \times 10 + \dfrac{1}{2}\times 0.5 \times {(10) }^{2}$

$\sf\implies s = 100 + \dfrac{1}{2}\times 0.5 \times 100$

$\sf\implies s = 100 + (0.5 \times 50)$

$\sf\implies s = 100 + 25$

$\sf\implies s = 125$

$\underline{\boxed{\purple{ \sf \therefore Distance = 125m}}}$

Answer 2

Given :-

• Initial velocity = 36 km/h

• Final velocity = 54 km/h

• Time = 10 s

To Find :-

• Acceleration of the car

• Distance travelled by car

Solution :-

By using First law of motion

$\implies \underline{\boxed{ \bf v = u + at}} \\ \\ \implies \sf 15 = 10 + a \times 10 \\ \\ \implies \sf 15 - 10 = 10a \\ \\\implies \sf 10a = 5 \\ \\\implies \sf a = \frac{5}{10} \\ \\\implies\underline{ \boxed{ \sf a = 0.5 \: {ms}^{ - 2}}}$

Now , By using 2nd law of motion

$\implies \underline{\boxed{ \bf s = ut + \frac{1}{2} {at}^{2}}} \\ \\ \implies \sf s = 10 \times 10 + \frac{1}{2} \times 0.5 \times { (10)}^{2} \\ \\ \implies \sf s =100 + \frac{1}{2} \times 50 \\ \\ \implies \sf s =100 + 25 \\ \\ \implies \underline{ \boxed{\sf s =125 \: m}}$