Question

A racing car has a uniform acceleration of 6 m/s^2.What distance Will it cover in 12 s after start?

Answer 1

Required Solution:

Given:

• Acceleration (a) = 6m/s²

• Initial velocity (u) = 0m/s

• time taken (t) = 12s

To calculate:

• Distance covered (s)

Calculation:

By applying the physical quantities in the third equation of motion:

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

$\sf { ⇢ s = (0 \times 12) + \dfrac{1}{\cancel{2}} \times \cancel{6} \times {12}^{2} }$

$\sf { ⇢ s = 3 \times 144 }$

$\boxed{ \sf { ⇢ s = 432 m}}$

Therefore, distance covered by the car is 432 m.

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More information:

Equations of motion:

• v = u +at
• s = ut + ½at²
• v² - u² = 2as

Remember!

• When the body starts from rest, it's initial velocity is 0.

• When body comes to rest or apply brakes, it's final velocity is 0.

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Answer 2

Answer: S= 432 m

Explanation:

• The equation of the motion is $S=ut+\frac{1}{2}at^{2}$
• Given, a=6 $m/s^{2}$ , u=0 and t=12 s.
• S= 0+$\frac{1}{2}(6)(12)^{2}$

         = 3*144

         = 432 m

• Hence, the displacement of a body starting from rest at an acceleration of 6  $m/s^{2}$  for 12 seconds is 432 meters.

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