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torisob [31]
3 years ago
11

Un automovil parte del reposo y acelera uniformemente hasta alcanzar una rapidez de 0,255km/h en un tiempo de 3/4 Minutos determ

inar su aceleracion y la distancia recorrida...
Ayuda pliis :'c
Physics
1 answer:
Elden [556K]3 years ago
5 0

Answer:

a = 1.5*10^-3 m/s^2

x = 0.033m = 3.3cm

Explanation:

To calculate the acceleration and the distance traveled by the car you use the following formulas:

v=v_o+at    (1)

x=v_ot+\frac{1}{2}at^2   (2)

v: final velocity = 0,255 km/h

vo: initial velocity = 0 m/s

t: time = 3/4 min

a: acceleration = ?

x: distance

In order to use the equations (1) and (2) you first convert the units of the final velocity to m/s, and the time to seconds.

v=0,255\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}\\\\v=0.07m/s\\\\t=\frac{3}{4}min*\frac{60s}{1min}=45s

Next, you solve the equation (1) for the acceleration a:

a=\frac{v}{t}=\frac{0.07m/s}{45s}=1.5*10^{-3}\frac{m}{s^2}

With this value of a you can calculate the distance traveled by the car, by using the equation (2):

x=\frac{1}{2}(1.5*10^{-3}m/s^2)(45s)^2=0.033m=3.3cm

hence, the acceleration of the car is 1.5*10^-3 m/s^2 and the distance traveled in 3/4 min is 0.033m

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