Question #38b0a

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Answer 1 · 2018-02-01T12:58:16

The acceleration is $=5.49ms^-2$

Apply the equation of motion

$s=ut+1/2at^2$

The initial velocity is $u=0ms^-1$

The time is $t=6.75s$

The distance is $s=125m$

The acceleration is

$a=2((s-ut)/t^2)$

$=2*((125-0)/6.75^2)$

$=5.49ms^-2$

Answer 2 · 2018-02-01T13:00:15

$5.49 m/(s^2)$

$s=v*t +½at^2$ . But v=0. Rearrange formula.

$a=(2s)/t^2$

$a=(2*125m)/(6.75s)^2=(250m)/(6.25 s)^2=(250m)/(45.56s^2)=5.49m/s^2$

Answer 3 · 2018-02-01T13:02:16

$"Acceleration"~~5.49ms^(-2)$

$s=ut+1/2at^2$ , where:

$s=125m$
$u=0ms^(-1)$
$t=6.75s$

$125=0(6.75)+a(6.75)^2/2$
$color(white)(125)=0+22.78125a$

$a=125/22.78125~~5.49ms^(-2)$