Answer :
Average acceleration, \(a=-7.71\times 10^{-5}\ m/s^2\)
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, \(a=\dfrac{v-u}{t}\)
\(a=\dfrac{0-26\ km/h}{0.36\ h}\)
\(a=-72.2\ km/h^2=-7.71\times 10^{-5}\ m/s^2\)
So, the magnitude of ship's acceleration is \(a=-7.71\times 10^{-5}\ m/s^2\).
(b) Average velocity, \(v_a=\dfrac{v+u}{2}\)
\(v_a=\dfrac{26\ m/s+0}{2}\)
\(v_a=13\ km/h\)
So, ship's average velocity is 13 km/h.
Hence, this is the required solution.
