Question

A particle from a point with a velocity of 6m/s and moves wuth an acceleration of -2m/s square . show that after 6s the particle will be at the starting point.

Answer 1

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

$\green{\underline \bold{Given:}} \\ \tt: \implies Initial \: velocity(u) = 6 \: m/s \\ \\ \tt: \implies Acceleration(a) = - 2 \: {m/s}^{2} \\ \\ \tt: \implies Time(t) = 6 \: sec \\ \\ \\ \red{\underline \bold{To \: Show:}} \\ \tt: \implies After \: 6 \: sec \: particle \: will \: be \: on \: starting \: point$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies s = ut + \frac{1}{2}a {t}^{2} \\ \\ \tt: \implies s = 6 \times 6 + \frac{1}{2} \times ( - 2) \times {6}^{2} \\ \\ \tt: \implies s = 36 - {6}^{2} \\ \\ \tt: \implies s =36 - 36 \\ \\ \tt: \implies s =0 \: m \\ \\ \green{\tt \therefore Displacement \: is \: 0 \: m} \\ \: \green{\tt So, \: after \: 6 \: sec \: particle \: will \: be \: on \: starting \: point} \\ \\ \blue{\bold{Some \: related \: formula}} \\ \orange{\tt \circ \:v = u + at} \\ \\ \orange{\tt \circ \: {v}^{2} = {u}^{2} + 2as}$

Answer 2

Given

• Initial velocity (u) = 6 m/s

• Acceleration (a) = -2 m/s²

• Time(t) = 6 seconds.

To ShoW

• The particle will be in its initial or starting point after 6 seconds.

Solution

[Note : After 6 seconds, the particle will be in its initial point means that the particle didn't cross any distance after 6 seconds ]

As we know equation of kinematics :

◗ s = ut + ½ at²

Putting values :

⇒ s = (6 × 6) + ½ (-2 × 6²)

= 36 + ½ (-2 × 36)

= 36 + ½ (-72)

= 36 - 36

= 0

As here, distance travelled (s) = 0, so that means the particle didn't cross any distance and so the distrance will be in its starting point.

Therefore,

It's showed that after 6 seconds the particle will be in its starting point.