1. The document contains two questions about velocity and acceleration. The first asks about the average velocity of a car traveling north for 3/4 of the time at 33 m/s and south for 1/4 of the time at 27 m/s. The average velocity is calculated to be 18 m/s. 2. The second question asks about an emu running north at 13 m/s and slowing to 9.4 m/s over 4.8 seconds. Its acceleration is calculated to be -0.75 m/s^2, and after an additional 2.7 seconds its velocity is calculated to be 11 m/s.

1. Please answer both questions:
1. A car makes a trip due north for three-fourths of the time and due south one-fourth of the time.
The average northward velocity has a magnitude of 33 m/s, and the average southward velocity
has a magnitude of 27 m/s. Taking northward to be the positive direction, what is the average
velocity for the trip?
2.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 9.40 m/s in 4.80 s. (a) What is the bird’s acceleration? Assume that
direction north is positive and direction south is negative. (b) Assuming that the acceleration
remains the same, what is the bird’s velocity after an additional 2.70 s has elapsed?
Solution
here,
1)
let the total time taken be t
the average velocity , v = total displacement /total time taken
v = ( (3t/4) * 33 - (t/4) * 27) /( t)
v = 18 m/s
2)
intial speed , u = 13 m/s
final speed , v = 9.4 m/s
time taken , t = 4.8 s
a)

2. let the bird's accelration be a
v = u + a * t
9.4 = 13 + a * 4.8
solving for a
a = - 0.75 m/s^2
b)
at t = 2.7 s
the bird's velocity , v = u + a * t
v = 13 - 0.75 * 2.7 m/s
v = 11 m/s