Question

A skateboarder is moving at 1.75 m/s when she starts going up an incline that causes an acceleration of -0.20 m/s2

How far does she go before she comes to a stop?

Answer 1

Approximately
`7.66\; \rm m`.

Step-by-step explanation:

Solve this question with a speed-time plot

The skateboarder started with an initial speed of
`u = 1.75\; \rm m \cdot s^(-1)` and came to a stop when her speed became
`v = 0\; \rm m \cdot s^(-1)`. How much time would that take if her acceleration is
`a = -0.20\; \rm m \cdot s^(-1)`?

`\begin{aligned} t &= (v - u)/(a) \\ &= (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}`.

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:

`\displaystyle (1)/(2) * 1.75\; \rm m \cdot s^(-1) * 8.75\; \rm s \approx 7.66\; \rm m`.

In other words, the skateboarder travelled
`15.3\; \rm m` 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:

`\displaystyle x = (1)/(2)\, (u + v) \, t = (1)/(2)\, (u + v)\cdot (v - u)/(a)= (v^2 - u^2)/(2\, a)`,

where:

• 
  `u` and
  `v` are the initial and final velocity of the object,
• 
  `a` is the constant acceleration that changed the velocity of this object from
  `u` to
  `v`, and
• 
  `x` is the distance that this object travelled while its velocity changed from
  `u` to
  `v`.

For the skateboarder in this question:

`\begin{aligned}x &= (v^2 - u^2)/(2\, a)\\ &= (\left(0\; \rm m \cdot s^(-1)\right)^2 - \left(1.75\; \rm m \cdot s^(-1)\right)^2)/(2* \left(-0.20\; \rm m \cdot s^(-2)\right))\approx 7.66\; \rm m \end{aligned}`.