3. En una carrera cuyo recorrido es recto, una moto circula durante 30 segundos hasta alcanzar una velocidad de 162.00km/h. Si la aceleración sigue siendo la misma, ¿cuánto tiempo tardará en recorrer los 200 metros que faltan para rebasar la meta y a qué velocidad lo hará?

4s50m/s (straight)

Explanation:

I will answer in English.

The translated question is:

3. In a race whose route is straight, a motorcycle circulates for 30 seconds until it reaches a speed of 162.00km / h. If the acceleration remains the same, how long will it take to travel the remaining 200 meters to pass the finish line and at what speed will it do so?Solution

1. Time

First, you must calculate the acceleration, a,  starting from rest (V₀ = 0), with a time fo 30 seconds, and final veloficty, V₁, of 162.00km/h.

Convert 162.00km/h to m/s:

162.00km/h × 1,000m/km × 3600s/h = 45m/sa = (V₁ - V₀)/t = (45m/s - 0) / (30s) = 1.5m/s²

Now, you can calculate the time to travel 200 meters

The most direct equation is;

d = V₁·t + a × t² / 2

Where you know:

V₁ = 45m/sa = 1.5m/s²d = 200m200 = 45t + 0.75t²0.75t² + 45t - 200 = 0

Use the quadratic formula with a = 0.75, b = 45, and c = -200

   

Discard the negative value. The positive value is t = 4.1565s

Round to one significant figure: 4s ← answer

2. Velocity

The speed or magnitude of velocity may be calculated using the equation:

V₂ = V₁ + a × t

With:

V₁ = 45m/sa = 1.5m/s²t = 4.1565sV₂ = 45m/s + 1.5m/s² × 4.1565s = 51m/s ≈ 50m/s, along the same straight line ← answer

B) raising the temperature and reducing the volume of the gas.|-/ 

Gas will most likely  fill the container