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 10.0 m/s in 3.30 s. (a) What is the magnitude and direction of the bird’s acceleration? (b) Assuming that the acceleration remains the same, what is the bird’s velocity after an additional 1.20 s has elapsed?

(a)

Magnitude of Acceleration = 0.9 m/s²

Direction of Acceleration = South

(b)

Vf = 8.9 m/s

Explanation:

(a)

The acceleration is given by the following formula:

a = (Vf - Vi)/t

where,

a = acceleration = ?

Vf = Final Velocity = 10 m/s

Vi = Initial Velocity = 13 m/s

t = time taken = 3.3 s

Therefore,

a = (10 m/s - 13 m/s)/3.3 s

a = - 0.9 m/s²

Here, negative sign shows that the direction of acceleration is opposite to the direction of motion of the bird. Therefore, direction of acceleration will be due South.

Magnitude of Acceleration = 0.9 m/s²

Direction of Acceleration = South

(b)

a = (Vf - Vi)/t

where,

a = acceleration = - 0.9 m/s²

Vf = Final Velocity = ?

Vi = Initial Velocity = 10 m/s

t = time taken = 1.2 s

Therefore,

- 0.9 m/s² = (Vf - 10 m/s)/1.2 s

(- 0.9 m/s²)(1.2 s) + 10 m/s = Vf

Vf = 8.9 m/s

0.3x0.21x0.08kg=mass 1 sheet

mass 500 = 500xmass 1

volume pile = 0.3x0.2x0.048

density=mass/volume= 500xmass 1/0.3x0.2x0.048

v patient student

the answer is to check all of them! they all apply well at least i think so!