Question

A speedboat moving at 29.0 m/s approaches a no-wake buoy marker 100 m ahead. The pilot slows the boat with a constant acceleration of -3.4 m/s2 by reducing the throttle. (a) How long does it take the boat to reach the buoy

Answer 1

`t =4.8sec`

Step-by-step explanation:

From the question we are told that

Velocity of speed boat
`V=29.0m/s`

Distance to Marker
`d=100`

Acceleration of
`a=-3.4m/s^2`

Generally the Newtons 3rd motion equation is given as

`v^2 = u^2 + 2 * a* s`

`v^2 = 29^2 + 2 * -3.4* 100`

`v = √(161)`

`v=12.68m/s`

Generally the Newton's first equation of motion is given as

`v = u + a*t`

`12.68 = 29 -3.4*t`

`12.68-29 = -3.4t`

`-16.32 = -3.4t`

`t =(-16.32)/(-3.4)`

`t =4.8sec` .