# two objects, x and y, accelerate from rest with the same constant acceleration. object x accelerates for twice the time as object y. which of the following is true of these objects at the end of their respective periods of acceleration?

### James

Guys, does anyone know the answer?

get two objects, x and y, accelerate from rest with the same constant acceleration. object x accelerates for twice the time as object y. which of the following is true of these objects at the end of their respective periods of acceleration? from EN Bilgi.

## Answer in Mechanics

Let speed of object X be ${v}_{x}$vx

Let speed of object Y be ${v}_{y}$vy

Let acceleration be $a$a

Let time of acceleration of object X be ${t}_{x}$tx

Let time of acceleration of object Y be ${t}_{y}$ty

${v}_{y}=0+a{t}_{y}$vy=0+aty

${t}_{y}=\frac{{v}_{y}}{a}\text{}\left(1\right)$ty=avy (1)

${v}_{x}=0+a{t}_{x}$vx=0+atx

${t}_{x}=\frac{{v}_{x}}{a}\text{}\left(2\right)$tx=avx (2)

${t}_{x}=3{t}_{y}\text{}\left(3\right)$tx=3ty (3)

from equation (1),(2) and (3)

${v}_{x}=3{v}_{y}$vx=3vy

$\therefore $∴ ** The final speed of object X is three times faster than object Y**

${v}_{x}^{2}=2a{s}_{x}$vx2=2asx

${v}_{y}^{2}=2a{s}_{y}$vy2=2asy

${v}_{x}^{2}=9{v}_{y}^{2}$vx2=9vy2

$\therefore \text{}{s}_{x}=9{s}_{y}$∴ sx=9sy

**Object X has travelled nine times as far as object Y**

Source : **www.assignmentexpert.com**

## Accelerate for twice the time

Two objects, A and B, are accelerated from rest with the same constant acceleration. However, object A is accelerated for twice as much time as object B. Which one of the following statements is true?

Source : **media.ed.science.psu.edu**

## Objects A and B both start from rest. They both accelerate at the same rate. However object A accelerates for twice the time as object B. Now, the distance traveled by object A compared to that of object B is four times as far. How was this derived?

The kinematic equation that works best for this is:

s = ut + 1/2 a t^2

63 = 12 t + 1/2 (6) t^2

63 = 12 t + 3 t^2

3 t^2 +12 t - 63 = 0

Divide each side by 3…

t^2 +4 t -21 = 0

Factoring…

(t+7)(t-3) = 0

So, either

t+7 = 0

t -3 = 0

Therefore

Either

t = -7

t = 3

Choose t = 3 for the only realistic answer for this situation.

It takes 3 seconds to go 63 meters.

Guys, does anyone know the answer?