Question

a car initially at rest starts moving with constant acceleration of 0.5m/second square and travels a distance of 25m. find its final velocity and time taken

Answer 1

Hiii...
Here is your answer..

Solution:-
Initial velocity=0
Acceleration= .5m/s^2
Distance travelled=25m
Final velocity=?

By using equation:-
$= > {v}^{2} = {u}^{2} + 2as \\ = > {v}^{2} = {(0)}^{2} + 2 \times .5 \times 25 \\ = > {v}^{2} = 25 \\ = > {v}^{2} = {5}^{2}$
$= > v = 5$
Now,
Final velocity=5m/s

We have to find time now,we will find it by using equation:-
$= > v = u + at \\ = > 5 = 0 + .5 \times t \\ = > 5 = .5t \\ = > t = \frac{5}{.5}$
$= > t = 10$
Hence,time =10 seconds

Hope it helps uh...✌️✌️✌️

Answer 2

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

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

$\green{\therefore{\text{Time\:taken=10\:sec}}}$

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

• In the given question information given about a car initially at rest starts moving with constant acceleration of 0.5m/second square and travels a distance of 25m.

• We have to find its final velocity and time taken.

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

• According to given question :

$\bold{Using \: second \: equation \: of \: motion } \\ : \implies s = ut + \frac{1}{2} {at}^{2} \\ \\ : \implies 25 = 0 \times t + \frac{1}{2} \times \frac{1}{2}\times {t}^{2} \\ \\ : \implies 25\times 4= t^{2}\\ \\ : \implies t =\sqrt{100} \\ \\ \green{ : \implies \text{t =10\:sec}}$

$\bold{using \: first \: equation \: of \: motion} \\ : \implies v = u + at \\ \\ : \implies v = 0 + 0.5 \times 10 \\ \\ \green{: \implies \text{v = 5 \: m/s}}$