An alligator swims to the left with a constant velocity of 5 \,\dfrac{\text{m}}{\text s}5 s m 5, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction. When the alligator sees a bird straight ahead, the alligator speeds up with a constant acceleration of 3 \,\dfrac{\text {m}}{\text s^2}3 s 2 m 3, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction leftward until it reaches a final velocity of 35 \,\dfrac{\text {m}}{\text s}35 s m 35, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction leftward. How many seconds does it take the alligator to speed up from 5 \,\dfrac{\text {m}}{\text s}5 s m 5, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction to 35 \,\dfrac{\text {m}}{\text s}35 s m 35, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction?
Answers
Answer:
10 sec
Explanation:
The equation for acceleration a is:
a= final velocity -initial velocity
time
Rearranging the equation to solve for the time interval \Delta tΔtdelta, t gives
We can calculate the time interval from the final velocity v_fv
f
v, start subscript, f, end subscript, initial velocity v_iv
i
v, start subscript, i, end subscript, and acceleration aaa.
\begin{aligned}\Delta t&=\dfrac{v_f-v_i}{a}\\ \\ &=\dfrac{-35\,\dfrac{\text {m}}{\text s}-\left(-5\,\dfrac{\text {m}}{\text s}\right)}{-3\,\dfrac{\text {m}}{\text s^2}}\\ \\ &=10\,\text s\end{aligned}
Δt
=
a
v
f
−v
i
=
−3
s
2
m
−35
s
m
−(−5
s
m
)
=10s
Hint #33 / 3
It takes the alligator 10\,\text s10s10, start text, s, end text to speed up from -5 \,\dfrac{\text {m}}{\text s}−5
s
m
minus, 5, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction to -35 \,\dfrac{\text {m}}{\text s}−35
s
m
minus, 35, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction.