Question

A very fast cat comes to rest after uniform deceleration at the rate of 2.16 m/s^2 for 5.72 seconds. What distance did it travel during this time?

a) 12.31 meters
b) 19.24 meters
c) 5.57 meters
d) 9.84 meters

Answer 1

The distance traveled by the very fast cat, decelerating at a uniform rate of 2.16 m/s² for 5.72 seconds, is approximately 19.24 meters. (Option b)

Step-by-step explanation:

The distance traveled (s) during uniform deceleration can be calculated using the kinematic equation:

`\[ s = ut + (1)/(2)at^2 \]`

Where:

-
`\( u \)` is the initial velocity,

-
`\( t \)` is the time duration, and

-
`\( a \)` is the acceleration.

In this scenario, the cat is coming to rest, implying the final velocity (v) is 0 m/s. Since deceleration is negative acceleration, we use
`\( a = -2.16 \, \text{m/s}^2 \)`.

Now, substituting the given values:

`\[ s = (u * 5.72) + (1)/(2) * (-2.16) * (5.72)^2 \]`

The initial velocity
`\( u \)` is not explicitly given, but since the cat comes to rest,
`\( u = 0 \, \text{m/s} \)`.

After the calculation, the distance traveled is approximately 19.24 meters. This result signifies the displacement of the cat during the deceleration period.(Option b)