A skateboarder starts from rest at the top of a straight ramp and accelerates uniformly down the ramp at 0.4 m s^-2. Find the distance she has travelled after 5 seconds.

AS Level Maths 9709: Distance from Rest Under Constant Acceleration

Step 1: Identify the suvat quantities

The skateboarder starts from rest, so $u=0\text{ m s}^{-1}$ . She accelerates uniformly, so we can use the suvat equation linking displacement, initial velocity, acceleration and time:

$s = ut + \frac12 at^2$

Here $u=0\text{ m s}^{-1}$ , $a=0.4\text{ m s}^{-2}$ , $t=5\text{ s}$ .

Step 2: Substitute and calculate

$s = (0)(5) + \frac12(0.4)(5)^2 = 0 + \frac12(0.4)(25)$

$s = 0.2 \times 25 = 5$

So $s = 5\text{ m}$ .

Step 3: Recompute independently as a check

Working the multiplication in a different order: $0.4\times 25 = 10$ , and then $\tfrac12\times10 = 5$ . This matches Step 2 exactly, confirming $s=5\text{ m}$ .

Why the other options are wrong

B ( $10\text{ m}$ ): comes from omitting the factor of $\tfrac12$ , i.e. computing $at^2=0.4\times25=10$ .
C ( $2\text{ m}$ ): is the final velocity $v=u+at=0+0.4\times5=2\text{ m s}^{-1}$ , mistaken for a distance.
D ( $1\text{ m}$ ): comes from using $t$ instead of $t^2$ , i.e. $\tfrac12\times0.4\times5=1$ .

Final answer

$\boxed{s = 5\text{ m}} \quad \text{(Option A)}$