Question

A tiger starts at rest and accelerates at a rate of 6.7 m/s². The tiger eventually reaches a top speed of 17 m/s. What is the tiger's displacement?

Answer 1

Approximately
`22\; {\rm m}`.

Step-by-step explanation:

In this question, the following information about the motion are given:

• Velocity before the acceleration:
  `u = 0\; {\rm m\cdot s^(-1)}`, since the tiger started at rest.
• Velocity after the acceleration:
  `v = 17\; {\rm m\cdot s^(-1)}`.
• Acceleration:
  `a = 6.7\; {\rm m\cdot s^(-2)}`.

The displacement
`x` during the acceleration can be found with the following SUVAT equation:

`\begin{aligned}x &= (v^(2) - u^(2))/(2\, a) \\ &= ((17)^(2) - (0)^(2))/(2\, (6.7))\; {\rm m}\\ &\approx 22\; {\rm m}\end{aligned}`.

In other words, the displacement during this motion would be approximately
`22 {\rm m}`.

There are other SUVAT equations that can also be used to solve this problem. However, this particular SUVAT equation
`x = (v^(2) - u^(2)) / (2\, a)` allows the answer to be reached in one step. The motivation behind this choice is that this SUVAT equation does not include the duration
`t` of the motion- which is neither provided nor required in this question.