Answer:(a) Choose the initial point where the pilot reduces the throttle and the final point where the boat passes the buoy: X
i
=0,X
f
=100m,
V
xi
=30m/s,V
xf
=?,a
x
=−3.5m/s
2
,andt=?
X
f
=X
i
+V
xi
t+
2
1
a
x
t
2
100m=0+(30m/s)t+
2
1
(−3.5m/s
2
)t
2
(1.75m/s
2
)t
2
−(30m/s)t+100m=0
We use the quadratic formula:
t=
2a
−b±
b
2
−4ac
t=
2(1.75m/s
2
)
30m/s±
900m
2
/s
2
−4(1.75m/s
2
)(100m)
=
3.5m/s
2
30m/s±14.1m/s
=12.6s or 4.53s
The smaller value is the physical answer. If the boat kept moving with the same acceleration, it would stop and move backward, then gain speed, and pass the buoy again at 12.6s.
(b) V
xf
=V
xi
+a
x
t=30m/s−(3.5m/s
2
)4.53s=14.1m/s
Explanation: