How would you calculate the speed (in $\frac{m}{s}$ $m/s$ ) of a skateboarder who accelerates from rest for 4 seconds down a ramp at an acceleration of $3 \frac{m}{s^{2}}$ $3 m/s^2$ ?

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Answer 1 · 2018-04-02T10:35:14

Applicable kinematic expression is

$v = u + a t$ $v=u+at$

Inserting given values we get

$v = 0 + 3 \times 4$ $v=0+3xx4$
$v=12\ ms^-1$

Answer 2 · 2018-04-02T10:59:51

$12 \ "m/s"$

Using the basic kinematics equation,

$v=u+at$

$u$ is the initial speed

$a$ is the acceleration

$t$ is the time

And so,

$v=0 \ "m/s"+3 \ "m/s"^2*4 \ "s"$

$=12 \ "m/s"$

How would you calculate the speed (in m/smsm/s) of a skateboarder who accelerates from rest for 4 seconds down a ramp at an acceleration of 3 m/s^23ms23 m/s^2?