Question

A car moving at 5.35 m/s picks up speed at a rate of 4.25 m/s for 34.6 m. What is the final velocity ?​

Answer 1

Given :-

☞$\:\begin{cases}\sf\red{Initial\: velocity \:(u)\: =\: 5.35m/s}\\\sf\orange{Acceleration\: ( a ) \:= \:4.25m/s^2 }\\\sf\purple{Distance\: (s) \ = \ 34.6m} \end{cases}$

To find :-

• $\bf\green{Final \ velocity}$ of the car

Solution :-

Using 3rd equation of motion

$\to\:\:\sf{v^2 = u^2 +2as }$

$\to\:\:\sf{v^2 = (5.35)^2 + 2\times 4.25 \times 34.6 }$

$\to\:\:\sf{ v^2 = 28.6225 + 8.50\times 34.6}$

$\to\:\:\sf{ v^2 = 28.6225 + 294.1}$

$\to\:\:\sf{v^2 = 322.7225}$

$\to\:\:\sf{ v = \sqrt{322.7225}}$

$\to\:\:\sf{v = 18m/s }$

$\therefore\:\sf The\: final\: velocity\: of\: the \:car \:is\: {\bf{\red{18m/s}}}$

Additional Information :-

Few more equations of motion

• $\sf\red{v = u + at}$

• $\sf\orange{ s = \dfrac{1}{2} (u + v) t}$

• $\sf\purple{s = ut + \dfrac{1}{2} \:at^2}$