Question

A car starting from rest attains a velocity of 36m/s with an accelaration of 4m/s².calculate the distance travelled by the car in 12 Sec

Answer 1

a=4m/s^2

t=12 sec

v=36m/s

u=0m/s

s=?

v^2-u^2=2as

36×36/2×4=s

s=162 m

Answer 2

GIVEN:

• Initial Velocity (u) = 0
• Final velocity (v) = 36 m/s
• Acceleration (a) = 4 m/s²
• Time taken (t) = 12 seconds

TO FIND:

• What is the distance travelled by the car in 12 seconds ?

SOLUTION:

Let the distance travelled by the car be 's' m

To find the distance travelled by the car, we use the 2) equation of motion

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

According to question:-

On putting the given values in the formula, we get

$\sf{\rightarrow s = 0 \times 12 + \dfrac{1}{2} \times 4 \times (12)^2}$

$\sf{\rightarrow s = 0 + \dfrac{1}{\cancel{2}} \times \cancel{4} \times (12)^2}$

$\sf{\rightarrow s = 2 \times 144}$

$\bf{\rightarrow \star \: s = 288 \: m \: \star}$

❝ Hence, the distance travelled by the car is 288 m ❞

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✫ Extra Information ✫

➨ First equation of motion = v = u + at

➨ Third Equation of motion = v² - u² = 2as