Question

A particle starts with a velocity of 5m/s and moves with a uniform acceleration of 2.5m/s2 its velocity after 4 sec is

Answer 1

Use the formula
v=u+at
where u=initial velocity
v=final velocity
a=acceleration
t =time

Therefore v=5+2.5×4
=5+10
=15m/s

Answer 2

Answer:

${\bold{\therefore Final\:velocity(v)=15\:m/s}}$

Explanation:

${\bold{\huge{\underline{SOLUTION-}}}}$

• In the given question information given about a particle which is moving in a straight line with velocity and uniform acceleration.

• We have to find the velocity of particle after 4 sec.

$\underline \bold{Given : } \\ \implies Initial \: velocity(u) = 5 \: ms \\ \implies Acceleration(a) = 2.5 \: m {s}^{2} \\ \implies Time(t) = 4 \: sec \\ \\ \underline \bold{To \: Find :} \\ \implies Final \: velocity(v) = ?$

• According to given question :

$\: \: \: \: \: \: \: \: \: \bold{First \: method }\\ \\ \bold{By \: First \: equation \: of \: motion : } \\ \implies v = u + at \\ \implies v = 5 + 2.5 \times 4 \\ \implies v = 5 + 10 \\ \bold {\implies v = 15 \: m/s}$

$\: \: \: \: \: \: \: \: \: \bold{Alternate \: method :} \\ \\ \bold{By \:Third \: equation \: of \: motion} \\ \implies s = ut + \frac{1}{2} a {t}^{2} \\ \implies s = 5 \times 4 + \frac{1}{2} \times 2.5 \times {4}^{2} \\ \implies s = 20 + \frac{1}{2} \times 2.5 \times 16 \\ \\ \implies s = 20 + 20 \\ \bold{\implies s = 40 \: m} \\ \\ \bold{by \: second \: equation \: of \: motion} \\ \implies {v}^{2} = {u}^{2} + 2as \\ \implies {v}^{2} = {5}^{2} + 2 \times 2.5 \times 40 \\ \implies {v}^{2} = 25 + 200 \\ \implies {v}^{2} = 225 \\ \implies v = \sqrt{225} \\ \bold{\implies v = 15 \: m/s}$