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

Answer:

$t =4.8sec$

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

 $v^2 = u^2 + 2 * a* s$

 $v^2 = 29^2 + 2 * -3.4* 100$

 $v = \sqrt{161}$

 $v=12.68m/s$

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

 $v = u + a*t$

 $12.68 = 29 -3.4*t$

 $12.68-29 = -3.4t$

 $-16.32 = -3.4t$

 $t =\frac{-16.32}{-3.4}$

 $t =4.8sec$ .