Question

a car travelling at velocity of 10 m/s due north speeds up uniformly to a velocity of 25 m/s in 3 s calculate the acceleration during this period.​

Answer 1

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

$\green{\tt{\therefore{Accceleration=5\:m/s^{2}}}}$

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

$\green{\underline \bold{Given :}} \\ \tt: {\implies Initial\: velocity(u) = 10 \: m/s} \\ \\ \tt : \implies Final\: velocity(v) = 25 \: m/s\\ \\ \tt: \implies Time(t) = 3\:sec \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Acceleration = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies v=u+at \\ \\ \tt: \implies 25= 10 + a\times 3 \\ \\ \tt: \implies 25- 10=a \times 3\\ \\ \tt: \implies 3a = 15 \\ \\ \tt:\implies a=\frac{15}{3} \\\\ \green{\tt: \implies a=5 \: m/s^{2}} \\ \\ \green{\tt\therefore Acceleration \: of\: car \: is \: 5 \: m/s^{2}} \\ \\ \blue{\bold {Some \: related \: formula}} \\ \orange{\tt \circ \: s = ut + \frac{1}{2}at^{2}} \\ \\ \orange{\tt \circ \: {v}^{2} = {u}^{2} +2as}$

Answer 2

Explanation:

Here,

V = u +at

a = (v-u)/t

a = (25 - 10)/3

a = 15/3

a = 5 m/s^2