Question

An Australian emu is running due north in a straight line at a
speed of 13.0 m/s and slows down to a speed of 10.6 m/s in 4.0 s.
(a) What is the direction of the bird’s acceleration? (b) Assuming that
the acceleration remains the same, what is the bird’s velocity after an
additional 2.0 s has elapsed?

Answer 1

Whats your idea on solving this? @PaaSolo

Answer 2

For part A, it says that the speed reduces. Originally the bird is running in north direction.
since the speed is reducing, the acceleration must be in opposite direction to that of the velocity. so what would be answer for part A?

Answer 3

direction of velocity is north. Velocity is reducing hence the acceleration must be in opposite direction. So the opposite direction to north is South, and that is the answer.

Answer 4

lets look at B First we need to find the value of acceleration Then use that acceleration to find speed after an additional 2 seconds.
Which formula do we use to get acceleration? are you familiar with Vf = Vi + at?

Answer 5

where Vf is final velocity, Vi is initial velocity
t is time, and a is acceleration.

Answer 6

You can go on from there, if you're familiar with the formula.

Answer 7

we have Vf = Vi + at
Vi = 13 m/s, Vf = 10.6 m/s, t = 4 s a is the unknown can you try to substitute those values and solve for a? 10.6 = 13 + a *4 I wrote it
That will give you:
10.6 - 13 = a *4
which means -1.4 = a*4 10.6 = 13 + a *4
I wrote it
That will give you:
10.6 - 13 = a *4
which means -1.4 = a*4

Answer 8

hence a = -0.35 m/s^2
now let us come to the real question. Once an additional 2 seconds pass, what is the new speed? Already the bird ran for 4 seconds.

Answer 9

After an additional 2 seconds, total time t= 6 s

Answer 10

to solve it, we go back to the same formula:Vf = Vi + at this time, use Vi = 13 m/s, a = -0.35 m/s^2, t= 6 s use those values to calculate Vf. Vf = 13 + (-0.35)*6
Vf = 13 -(0.35 *6)
Vf = 13 - (2.1)
Vf = 10.9 m/s

Answer 11

Hope I was helpful, good luck with your studies! [;

Answer 12

i got -0.6 not -0.35