1. ## displacement, velocity, acceleration

An object initially at rest accelerates 10ms^-2 over a distance of 20m, until it reaches a speed of 20m/s. Another object travelling at 25m/s accelerates at 8ms^-2 for the same time period. It is argued that the second object will travel a greater distance under these conditions, and object two will always travel the greater distance when in motion for the same time as object 1. Is this argument valid:

What I've done:

Object 1:

v(0) = 0m/s
a = 10ms^-2
s=20m
v = 20 m/s

Object 2
:

v(0) = 25
a = 8

Working:

Object one:

a = dv/dt = 10

v = 10t + c
v(0) = 0 , 0 = 10(0) +c, c = 0

ds/dt = 10t

s = 5t^2 + c

Object two:

a = dv/dt = 8

v = 8t+c
v(0) = 25, 25 = 8(0) + c , c =25
v = 8t + 25

ds/dt = 8t+25

s = 4t^2 +c

I am not sure what to do from here?

2. Hi Gul:

Your question should be on the physics subforum, or better in Physics Help Forum!.

In this problem both are movements with constant acceleration $a_1$ and $a_2$, given.

The equations you should know in order to solve: $v(t)=v_0+at$ and $v_{f}^{2}-v_{0}^2=2a \Delta x$.

With the first one you can compute the interval of time that corresponds to both movements, and with the second one the distance traveled by the object 2.