Question

An object starts from rest after a time it covers a displacement of 108 km. Calculate the time taken and the acceleration of the object of its has a velocity of 36 km/he

Answer 1

$\displaystyle\large\underline{\sf\red{Given}}$

✭ Initial Velocity (u) = 0 km/h

✭ Final Velocity (v) = 36 km/h

✭ Displacement (s) = 108 km

$\displaystyle\large\underline{\sf\blue{To \ Find}}$

◈ The time taken and the acceleration?

$\displaystyle\large\underline{\sf\gray{Solution}}$

So here we may first find the acceleration with the help of the third Equation of motion, that is,

$\displaystyle\sf \underline{\boxed{\sf v^2-u^2 = 2as}}$

And then find the time with the help of the first Equation of motion, that is,

$\displaystyle\sf \underline{\boxed{\sf v = u+at}}$

━━━━━━━━━

$\underline{\bigstar\:\textsf{According to the given Question :}}$

So now the acceleration will be,

➝ $\displaystyle\sf v^2-u^2 = 2as$

➝ $\displaystyle\sf 36^2+0^2 = 2\times a \times 108$

➝ $\displaystyle\sf 1296 = 216a$

➝ $\displaystyle\sf \dfrac{1296}{216} = a$

➝ $\displaystyle\sf \orange{Acceleration = 6 \ km/h^2}$

Now we shall find the time taken, that is,

➳ $\displaystyle\sf v = u+at$

➳ $\displaystyle\sf 36 = 0+6\times t$

➳ $\displaystyle\sf 36 = 6t$

➳ $\displaystyle\sf \dfrac{36}{6} = t$

➳ $\displaystyle\sf \pink{Time = 6 \ hours}$

$\displaystyle\underline{\sf\therefore Acc^n \ is \ 6km/h^2 \ \& \ Time \ is \ 6hr}$

━━━━━━━━━━━━━━━━━━

Answer 2

★ Given -:

The object starts from rest (u = 0).

displacement (s) = 108 km

velocity (v) = 36 km/hr

★ To find -:

The time by the body and the acceleration ..

★ Solution -:

We know that ,

v² = u² + 2as

by putting the values

(36)² = (0)² + 2 × a × 108

1296 = 216a

a = 1296/216

a = 6 km/hr²

________________________________

We know that,

v = u + at

by putting the values

36 = 0 + 6t

36 = 6t

t = 6 hrs