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. Assuming that the acceleration remains the same, what is the bird's velocity after an additional 2.0 s has elapsed?
v=u+ft or f=(v-u)/t=(10.6-13)/4=-2.4/4
=-0.6m/s^2.
Now the u is 10.6m/s and f=-0.6m/s^2
v=u+ft=10.6-0.6*2=10.6-1.2=9.4 m/s^2
To find the bird's velocity after an additional 2.0 s has elapsed, we need to calculate its final velocity using the given information.
Given:
Initial velocity (u) = 13.0 m/s
Final velocity (v) = 10.6 m/s
Time taken to slow down (t) = 4.0 s
Additional time passed (t₂) = 2.0 s
Using the formula for uniform acceleration:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
Since the bird is slowing down, the acceleration (a) can be calculated as:
a = (v - u) / t
Substituting the given values:
a = (10.6 - 13.0) / 4.0
a = -0.7 m/s²
Now, we can calculate the velocity after an additional 2.0 s has elapsed:
v₂ = u + a(t + t₂)
Substituting the known values:
v₂ = 13.0 + (-0.7)(4.0 + 2.0)
v₂ = 13.0 + (-0.7)(6.0)
v₂ = 13.0 - 4.2
v₂ = 8.8 m/s
Therefore, the bird's velocity after an additional 2.0 s has elapsed is 8.8 m/s.