Niamh is swimming in the sea. She sees an interesting fish ahead of her and speeds up to a velocity of 1 m/s. Her acceleration is 0.25 m/s2 and the change in velocity takes 2 seconds. Work out how fast Niamh was swimming before she saw the fish.
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Answer:
Question
A fish swimming in a horizontal plane has velocity
v
i
=(4.00
i
^
+1.00
j
^
)m/s at a point in the ocean where the position relative to a certain rock is
r
i
=(10.0
i
^
−4.00
j
^
)m. After the fish swims with constant acceleration for 20.0s, its velocity is
v
=(20.0
i
^
−5.00
j
^
)m/s. (a) What are the components of the acceleration of the fish? (b) What is the direction of its acceleration with respect to unit vector
i
^
? (c) If the fish maintains constant acceleration, where is it at t=25.0s and in what direction is it moving
Model the fish as a particle under constant acceleration. We use our old standard equations for constant-acceleration straight-line motion, with x and y subscripts to make them apply to parts of the whole motion. At t=0,
v
i
=(4.00
i
^
+1.00
j
^
)m/s and
r
^
i
=(10.00
i
^
−4.00
j
^
)m
At the first “final” point we consider, 20.0s later
v
f
=(20.0
i
^
−5.00
j
^
)m/s
(a) a
x
=
Δt
Δv
x
=
20.0s
20.0m/s−4.00m/s
=0.800m/s
2
a
y
=
Δt
Δv
y
=
20.0s
−5.00m/s−1.00m/s
=−0.300m/s
2
(b) θ=tan
−1
(
0.800m/s
2
−0.300m/s
2
)=−20.6
0
=339
0
(c) At t=25.0s the fish’s position is specified by its coordinates and the direction of its motion is specified by the direction angle of its velocity:
x
f
=x
i
+v
xi
t+
2
1
a
x
t
2
=10.0m+(4.00m/s)(25.0s)+
2
1
(0.800m/s
2
)(25.0s)
2
=360m
y
f
=y
i
+v
yi
t+
2
1
a
y
t
2
=−4.00m+(1.00m/s)(25.0s)+
2
1
(−0.300m/s
2
)(25.0s)
2
=72.7m
v
xf
=v
xi
+a
x
t=4.00m/s+(0.800m/s
2
)(25.0s)=24m/s
xf
=v
xi
+a
x
t=4.00m/s+(0.800m/s
2
)(25.0s)=24m/s
v
yf
=v
yi
+a
y
t=1.00m/s−(0.300m/s
2
)(25.0s)=−6.50m/s
θ=tan
−1
(
v
x
v
y
)=tan
−1
(
24.0m/s
−6.50m/s
)=−15.2
0.
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Answer:
0.5 m/s
Explanation:
The formula used to calculate acceleration (a) given the change in velocity (Δv) over a given time (t) is: a = Δv / t. Multiply both sides by t to get Δv as the subject: a t = Δv. All of the values in the question use standard units (m and s), so we can substitute in the values: Δv = 0.25 × 2 = 0.5 m/s. This is the change in Niamh's speed. Niamh's initial speed is: 1 – 0.5 = 0.5 m/s.