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:

Constant Acceleration Motion

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

The relation between the initial and final speeds is:

$v_f=v_o+a.t$

Where:

vf  = Final speed

vo = Initial speed

a   = Acceleration

t   = Elapsed time

The acceleration can be calculated by solving for a:

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

And the distance traveled is:

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

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

Let's convert the speeds to m/s. Recall 1 Km=1000 m and 1 h= 3600 s:

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

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

Calculating the acceleration:

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

$a = 0.28\ m/s^2$

Now for the distance:

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

$x=416.67\ m+125\ m$

$\boxed{x = 541.67\ m}$

The car will travel 541.67 m

Answer 2

The distance will be "541.67 m".

Uniform rate of acceleration:

According to the question,

• Final speed, $v_f = 80 \ km/h \ ,or \ 22.22 \ m/s$
• Initial speed, $v_o = 50 \ km/h, or \ 13.89 \ m/s$
• Elapsed time, $t = 30 \ s$

We know the relation,

→ $v_f = v_o +at$

or,

→   $a = \frac{v_f-v_o}{t}$

Now,

The acceleration,

→ $x = v_o.t+ \frac{a.t^2}{2}$

By substituting the values,

     $= \frac{22.22-12}{30}$

     $= 0.28 \ m/s^2$

hence,

The distance traveled be:

→ $x = 13.89\times 30+\frac{0.28.30^2}{2}$

      $= 416.67+125$

      $= 541.67 \ m$

Thus the above answer is right.

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https://brainly.com/question/4931057