What is the acceleration experienced by a car that takes 10 s to reach 27 m/s from rest?

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Answer 1 · 2018-05-21T00:45:04

$2.7 \frac{m}{s^{2}}$ $2.7 m/s^2$

Given: a car takes $10$ $10$ s to reach $27 \frac{m}{s}$ $27 m/s$ from rest.

acceleration $= a = \frac{v_{f} - v_{i}}{t}$ $= a = (v_f - v_i)/t$ ,

where $v_{f} =$ $v_f =$ velocity final and $v_{i} =$ $v_i =$ velocity initial

$a = \frac{27 - 0}{10} = 2.7 \frac{m}{s^{2}}$ $a = (27 - 0)/10 = 2.7 m/s^2$

Answer 2 · 2018-05-21T12:51:55

$2.7 \ "m/s"^2$

Acceleration is given through the equation,

$a=(v-u)/t$

where:

$v$ is the final velocity of the object

$u$ is the initial velocity of the object

$t$ is the time taken

Here, since the car was at rest, it was stationary, and so $u=0$ .

$:.a=v/t$

$=(27 \ "m/s")/(10 \ "s")$

$=2.7 \ "m/s"^2$