Question

How far does a car travel in 30.0 s while its velocity is changing from 50.0 km/h to 80.0 km/h at a uniform rate of acceleration?

Answer 1

Answer:

The car will travel 541.67 m

Explanation:

Uniform Acceleration

When an object changes its velocity at the same rate, the acceleration is constant.

The relation between the initial and final speeds is:

$v_f=v_o+a.t$

Where:

vf  = Final speed

vo = Initial speed

a   = Constant acceleration

t   = Elapsed time

The acceleration can be computed by solving for a:

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

And the distance traveled is calculated as follows:

$\displaystyle x=v_o.t+\frac{a.t^2}{2}$

The car travels during a time of t=30 s, and its speed changes from vo=50 Km/h to vf=80 Km/h.

Converting the speeds to m/s:

vo=50 Km/h * 1000/3600 = 13.89 m/s

vf=80 Km/h * 1000/3600 = 22.22 m/s

Both numbers are shown with a 2-decimal precision but they will be used with higher accuracy in further calculations.

Calculating the acceleration:

$\displaystyle a=\frac{22.22-13.89}{30}$

$a = 0.28\ m/s^2$

Now for the distance:

$\displaystyle x=13.89*30+\frac{0.28.30^2}{2}$

$x=416.67\ m+125\ m$

x = 541.67 m

The car will travel 541.67 m