Holly Guerrero

2022-01-06

An Australian emu is running due north in a straight line at a speed of $13.0\frac{m}{s}$ and slows down to a speed of $10.6\frac{m}{s}$ in 40 s.
a) What is the direction of the birds

boronganfh

(a) The emu is moving to the north with speed of ${\upsilon }_{0}=13m\cdot {s}^{-1}$ then it slows down to a speed of $\upsilon =10.6m\cdot {s}^{-1}$ during time interval of t = 4. The difference in the speeds is negative so according to:
$\stackrel{\to }{\upsilon }-\stackrel{\to }{{\upsilon }_{0}}=\stackrel{\to }{at}$
the acceleration must be negative (the time is always positive), so its in the opposite direction of the speeds, which
is to the north. And it has a magnitude of
$a=\frac{\stackrel{\to }{\upsilon }-\stackrel{\to }{{\upsilon }_{0}}}{t}$
$=\frac{\left(10.6m\cdot {s}^{-1}\right)-\left(13m\cdot {s}^{-1}\right)}{4s}$
$=-0.6m\cdot {s}^{-2}$
$a=-0.6m\cdot {s}^{-2}$
(b) Assume that the acceleration remains constant, we can get the final velocity after additional time of 2s, so $t=2+4=6s,$ using (1), the final velocity is therefore:
$\upsilon ={\upsilon }_{0}+at$
$=\left(13m\cdot {s}^{-1}\right)+\left(-0.6m\cdot {s}^{-2}\right)\left(6s\right)$
$=9.4m\cdot {s}^{-1}$
the final speed is therefore:
$\upsilon =9.4m\cdot {s}^{-1}$

Karen Robbins

The acceleration of the bird is the quotient obtained when the difference of its final and initial velocities is divided by the time elapsed.
$a=\frac{Vf-V1}{}$
$a=\frac{10.6\frac{m}{s}-13\frac{m}{s}}{4}=-0.6\frac{m}{{s}^{2}}$
Thus, the acceleration is equal to -0.6 $m/{s}^{2}$
. For the velocity after 2 seconds,
$V{f}_{2}=V{f}_{1}+at$
Substituting,
$Vf2=10.6\frac{m}{s}+\left(-0.6\right)×2=9.4\frac{m}{s}$

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