Question

A car accelerates from 36 Km/h to 54 Km/h in 10 seconds. Find acceleration and distance travelled by the car. (0.2 m/s2 , 125 m)

Answer 1

GIVEN:

• Initial Velocity (u) = 36 km/hr
• Final velocity (v) = 54 km/hr
• Time taken (t) = 10 Seconds

TO FIND:

• What is the acceleration and distance covered by the car ?

SOLUTION:

Let the acceleration of the car be 'a' m/s²

Convert km/hr into m/s

➜ Initial Velocity = 36 km/hr

➜  $\sf{ \cancel{36} \times \dfrac{5}{\cancel{18}}}$

➜  Initial Velocity = 10 m/s

➜ Final Velocity = 36 km/hr

➜ $\sf{ \cancel{54} \times \dfrac{5}{\cancel{18}}}$

➜ Final Velocity = 15 m/s

We know that the formula for finding the acceleration of the car is :-

$\bf{\star \: Acceleration = \dfrac{Final \: Velocity - Initial \: Velocity}{Time\: taken} \: \star}$

According to question:-

$\rm{\dashrightarrow a = \dfrac{15 - 10}{10} }$

$\rm{\dashrightarrow a = \cancel\dfrac{5}{10}}$

$\bf{\dashrightarrow \star \: a = 0.5 \: m/s^2\: \star}$

❝ The acceleration of the car is 0.5 m/s² ❞

We know that the formula for finding the distance covered by the car is:-

$\bf{\star \: s = ut + \dfrac{1}{2} at^2 \: \star}$

According to question:-

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

$\rm{\dashrightarrow s = 10 \times 10 + \dfrac{1}{\cancel{2}} \times 0.5 \times \cancel{100}}$

$\rm{\dashrightarrow s = 100 + 0.5 \times 50}$

$\rm{\dashrightarrow s = 100 + 25}$

$\bf{\dashrightarrow \star \: s = 125 \: m \: \star}$

❝ Hence, the distance covered by the car is 125 m and acceleration of the car is 0.5 m/s² ❞

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