Question

A second particle, Q, also moves along the x-axis so that its velocity for 0 £ £t 4 is given by Q t cos 0.063 ( ) t 2 v t( ) = 45 meters per hour. Find the time interval during which the velocity of particle Q is at least 60 meters per hour. Find the distance traveled by particle Q during the interval when the velocity of particle Q is at least 60 meters per hour.

Answer 1

The time interval when
`V_Q(t) \geq 60` is at
`1.866 \leq t \leq 3.519`

The distance is 106.109 m

Explanation:

The velocity of the second particle Q moving along the x-axis is :

`V_(Q)(t)=45√(t) cos(0.063 \ t^2)`

So ; the objective here is to find the time interval and the distance traveled by particle Q during the time interval.

We are also to that :

`V_Q(t) \geq 60` between
`0 \leq t \leq 4`

The schematic free body graphical representation of the above illustration was attached in the file below and the point when
`V_Q(t) \geq 60` is at 4 is obtained in the parabolic curve.

So,
`V_Q(t) \geq 60` is at
`1.866 \leq t \leq 3.519`

Taking the integral of the time interval in order to determine the distance; we have:

distance =
`\int\limits^(3.519)_(1.866) {V_Q(t)} \, dt`

=
`\int\limits^(3.519)_(1.866) {45√(t) cos(0.063 \ t^2)} \, dt`

= By using the Scientific calculator notation;

distance = 106.109 m