Question

A sport car traveling eastward at 27.8 m/s slows at a constant rate to a stop in 8 minutes, what is the displacement of the sports car in this time interval?

Answer 1

`6672\; {\rm m}`.

Step-by-step explanation:

The question assumes that the velocity of this vehicle is changing at a constant rate, meaning that acceleration of this vehicle is constant. Displacement
`x` of this vehicle in the given time
`t` can be found with the SUVAT equation:

`\displaystyle x = \left((v + u)/(2)\right)\, t`,

Where:

• 
  `u = 27.8\; {\rm m\cdot s^(-1)}` is the initial velocity,
• 
  `v` is the velocity after the given time period (
  `v = 0\; {\rm m\cdot s^(-1)}` since the vehicle has stopped,) and
• 
  `t = 8\; \text{minute} = 640\; {\rm s}` is the time required for velocity to change from
  `u` to
  `v`.

Substitute in the value of
`u`,
`v`, and
`t` to find displacement
`x`:

`\begin{aligned} x &= \left((v + u)/(2)\right)\, t \\ &= \left((0 + 27.8)/(2)\right)\, (640)\; {\rm m} \\ &= 6672\; {\rm m}\end{aligned}`.

In other words, the distance travelled under the assumptions would be
`6672\; {\rm m}`.