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

Answer:

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

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:

$\displaystyle a=\frac{v_f-v_o}{t}$

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:

$\displaystyle a=\frac{12.5-20}{4}=\frac{-7.5}{4}$

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