When the velocity changes uniformly, the object has a constant acceleration. The acceleration, the velocities, and the distance are related by the equation:

$v_f^2=v_o^2+2ax$

Where:

vf = final velocity

vo = initial velocity

a = acceleration

x = distance

Solving for a:

$\displaystyle a=\frac{v_f^2-v_o^2}{2x}$

The car travels a distance of x=200 m and the velocities are:

vo = 216 Km/h

vf = 360 Km/h

Both velocities must be converted to meters by seconds.

vo = 216 Km/h *1000/3600 = 60 m/s

vf = 360 Km/h *1000/3600 = 100 m/s

The acceleration is:

$\displaystyle a=\frac{100^2-60^2}{2*200}$

$\displaystyle a=\frac{10000-3600}{400}$

$\displaystyle a=\frac{6400}{400}$

$\mathbf{a = 16\ m/s^2}$

7 0

3 years ago

You walk 45 m to the north, then turn 90° to your right and walk another 45 m How far are you from where you originally started?

docker41 [41]

C^2=a^2+b^2
c^2=45^2+45^2
c^2=4050
c=63.64
c=64
The answer is D.

3 0

4 years ago