Answer :
Acceleration = (change in speed) / (time for the change)
Change in speed= (0 - 26 km/hr) = -26 km/hr
(-26 km/hr) x (1,000 m/km) x (1 hr / 3,600 sec) = -7.222 m/sec
Average acceleration = (-7.222 m/s) / (22 min x 60sec/min) = -0.00547 m/sec²
Average speed during the stopping maneuver =
(1/2) (start speed + end speed) = 13 km/hr = 3.6111 m/sec
Chances of running aground:
If he stays in the designated shipping lanes, he ought to be OK.
Answer :
Average acceleration, [tex]a=-7.71\times 10^{-5}\ m/s^2[/tex]
Average velocity, v = 13 km/h
Explanation:
It is given that,
Initial velocity of ship, v = 26 Km/h
Final velocity of ship, u = 0
Distance covered by ship, d = 5 km
Time taken, t = 22 min = 0.36 h
(a) Average acceleration, [tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{0-26\ km/h}{0.36\ h}[/tex]
[tex]a=-72.2\ km/h^2=-7.71\times 10^{-5}\ m/s^2[/tex]
So, the magnitude of ship's acceleration is [tex]a=-7.71\times 10^{-5}\ m/s^2[/tex].
(b) Average velocity, [tex]v_a=\dfrac{v+u}{2}[/tex]
[tex]v_a=\dfrac{26\ m/s+0}{2}[/tex]
[tex]v_a=13\ km/h[/tex]
So, ship's average velocity is 13 km/h.
Hence, this is the required solution.
