Answer:
Approximately [tex]7.66\; \rm m[/tex].
Explanation:
Solve this question with a speed-time plot
The skateboarder started with an initial speed of [tex]u = 1.75\; \rm m \cdot s^{-1}[/tex] and came to a stop when her speed became [tex]v = 0\; \rm m \cdot s^{-1}[/tex]. How much time would that take if her acceleration is [tex]a = -0.20\; \rm m \cdot s^{-1}[/tex]?
[tex]\begin{aligned} t &= \frac{v - u}{a} \\ &= \frac{0\; \rm m \cdot s^{-1} - 1.75\; \rm m \cdot s^{-1}}{-0.20\; \rm m \cdot s^{-2}} \approx 8.75\; \rm s\end{aligned}[/tex].
Refer to the speed-time graph in the diagram attached. This diagram shows the velocity-time plot of this skateboarder between the time she reached the incline and the time when she came to a stop. This plot, along with the vertical speed axis and the horizontal time axis, form a triangle. The area of this triangle should be equal to the distance that the skateboarder travelled while she was moving up this incline until she came to a stop. For this particular question, that area is approximately equal to:
[tex]\displaystyle \frac{1}{2} \times 1.75\; \rm m \cdot s^{-1} \times 8.75\; \rm s \approx 7.66\; \rm m[/tex].
In other words, the skateboarder travelled [tex]15.3\; \rm m[/tex] up the slope until she came to a stop.
Solve this question with an SUVAT equation
A more general equation for this kind of motion is:
[tex]\displaystyle x = \frac{1}{2}\, (u + v) \, t = \frac{1}{2}\, (u + v)\cdot \frac{v - u}{a}= \frac{v^2 - u^2}{2\, a}[/tex],
where:
- [tex]u[/tex] and [tex]v[/tex] are the initial and final velocity of the object,
- [tex]a[/tex] is the constant acceleration that changed the velocity of this object from [tex]u[/tex] to [tex]v[/tex], and
- [tex]x[/tex] is the distance that this object travelled while its velocity changed from [tex]u[/tex] to [tex]v[/tex].
For the skateboarder in this question:
[tex]\begin{aligned}x &= \frac{v^2 - u^2}{2\, a}\\ &= \frac{\left(0\; \rm m \cdot s^{-1}\right)^2 - \left(1.75\; \rm m \cdot s^{-1}\right)^2}{2\times \left(-0.20\; \rm m \cdot s^{-2}\right)}\approx 7.66\; \rm m \end{aligned}[/tex].
