Question

If you are driving 72 Km/h along a straight road and you reduce your speed to 45 km/h in 4.0 s,what is your acceleration during this inattentive period?

Answer 1

`a=-1.875\ m/s^2`

Step-by-step explanation:

Motion With Constant Acceleration

It occurs when the velocity of an object changes uniformly in time.

The equation that rules the change of velocities is:

`v_f=v_o+at`

Where:

a = acceleration

vo = initial speed

vf = final speed

t = time

Solving for a:

`\displaystyle a=(v_f-v_o)/(t)`

The initial speed is vo=72 Km/h and it's reduced to vf=45 km/h in a time of t=4 s. Both speeds will be converted to m/s:

vo=72*1000/3600 = 20 m/s

vf=45*1000/3600 = 12.5 m/s

Now calculate the acceleration:

`\displaystyle a=(12.5-20)/(4)=(-7.5)/(4)`

`\mathbf{a=-1.875\ m/s^2}`