Question

A cyclist travelling at 8 ms-1 applies her brakes and comes to a stop in a distance of 5 metres. Calculate the acceleration of the cyclist.

Answer 1

`(-6.4)\; {\rm m\cdot s^(-2)}`, assuming that the acceleration of the cyclist is constant.

Step-by-step explanation:

For an object moving with a constant acceleration, the following SUVAT equation would apply:

`v^(2) - u^(2) = 2\, a\, x`,

Where:

• 
  `v` is the velocity after accelerating,
• 
  `u` is the velocity before accelerating,
• 
  `a` is acceleration, and
• 
  `x` is the distance travelled during this motion.

In this question, it is given that:

• 
  `v = 0\; {\rm m\cdot s^(-1)}` after the acceleration since the cyclist has stopped;
• 
  `u = 8\; {\rm m\cdot s^(-1)}` was the initial velocity before the acceleration,
• 
  `x = 5\; {\rm m}` is the distance travelled during the motion.

The value of acceleration
`a` needs to be found. Rearrange the equation
`v^(2) - u^(2) = 2\, a\, x` to find an expression for acceleration
`a\!`:

`\begin{aligned} a &= (v^(2) - u^(2))/(2\, x) \\ &= ((0)^(2) - (8)^(2))/(2\, (5))\; {\rm m\cdot s^(-2)} \\ &= (-6.4)\; {\rm m\cdot s^(-2)}\end{aligned}`.

In other words, the acceleration of this cyclist would be
`6.4\; {\rm m\cdot s^(-2)}`. Note that the value of acceleration is negative because the cyclist is slowing down.