Question

Times
c. A car initially at rest starts moving with a constant acceleration
of 0.5m/s2 and travels at a distance of 25m. Calculate the final
velocity.
121​

Answer 1

Answer & Explanation:

Use Torricelli's equation,

$v^2 = v_0^2 + 2 a d \, \, .$

Since the car starts from rest, $v_0 = 0$ and the final speed is

$v = \sqrt{ 2 a d } = \sqrt{ 2 \times ( 0.5 \, \text{m} / \text{s}^2) \times (25 \, \text{m}) } = 5 \, \text{m} / \text{s} \, \, .$

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{Final\:velocity=5\:m/s}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about car initially at rest starts moving with a constant acceleration

car initially at rest starts moving with a constant accelerationof 0.5m/s2 and travels at a distance of 25m.

• We have to find Final velocity of car.

$\green{\underline\bold{Given :}} \\ :\implies \text{Initial\:velocity(u)=0 m/s} \\ \\ :\implies \text{Acceleration(a)= 0.5\:m/s}^{2} \\ \\ :\implies \text{Distance(s)= 25\:m} \\ \\ \red{\underline \bold{To \: Find : }} \\ :\implies \text{Final\:velocity= ?}$

• According to given question :

$\bold{Using \: Third\: equation \: of \: motion : } \\ : \implies v^{2} = u^{2} + 2as\\ \\ : \implies v^{2}= 0 + 2\times 0.5\times 25 \\ \\ : \implies v^{2}= 25 \\\\ : \implies v= \sqrt{25} \\ \\ \green{ :\implies \text{v= 5 \: m/s}}$