Motion and Motion Graphs: Question 8

Worked solution

Step 1: Calculate the acceleration during the first phase

Acceleration is the change in velocity per unit time, $a = \dfrac{\Delta v}{\Delta t}$. During the first $4\text{ s}$, the speed changes from $2\text{ m/s}$ to $10\text{ m/s}$:

$a_1 = \frac{10\text{ m/s} - 2\text{ m/s}}{4\text{ s}} = \frac{8\text{ m/s}}{4\text{ s}} = 2.0\text{ m/s}^2$

Step 2: Calculate the acceleration while braking

During braking, the speed changes from $10\text{ m/s}$ to $0\text{ m/s}$ over $2\text{ s}$ (from $t = 4\text{ s}$ to $t = 6\text{ s}$):

$a_2 = \frac{0\text{ m/s} - 10\text{ m/s}}{2\text{ s}} = \frac{-10\text{ m/s}}{2\text{ s}} = -5.0\text{ m/s}^2$

The negative sign shows this is a deceleration. Its magnitude is $5.0\text{ m/s}^2$.

Step 3: Compare the two magnitudes

$5.0\text{ m/s}^2 > 2.0\text{ m/s}^2$

So the braking deceleration has the greater magnitude, the skateboarder slows down more sharply than she originally sped up.

Why the other options are wrong

Final answers

• Braking deceleration magnitude $=$ $5.0\text{ m/s}^2$, which is greater than the $2.0\text{ m/s}^2$ acceleration during the first $4\text{ s}$, option A