Respuesta :
Answer:
The acceleration and the distance are 25200 mi/h² and 0.1008 mi.
Explanation:
Given that,
Initial speed = 71 mi/h
Final speed = 50 mi/h
Time = 3.0 s
(a). We need to calculate the acceleration
Using equation of motion
[tex]v=u+at[/tex]
[tex]a=\dfrac{v-u}{t}[/tex]
Put the value in the equation
[tex]a=\dfrac{(50-71)\times3600}{3}[/tex]
[tex]a=-25200\ mi/h^2[/tex]
Negative sign shows the deceleration.
(b). We need to calculate the distance
Using equation of motion
[tex]v^2=u^2+2as[/tex]
[tex](50)^2=(71)^2+2\times(-25200)\times s[/tex]
[tex]s=\dfrac{(50)^2-(71)^2}{-25200}[/tex]
[tex]s=0.1008\ mi[/tex]
Hence, The acceleration and the distance are 25200 mi/h² and 0.1008 mi.
The distance traveled by the car when the car is constantly deaccelerating at a rate of 25200 miles/h² is 0.0504 miles.
Given to us
Initial Velocity of the car, u = 71 miles/h
Final Velocity of the car, v = 50 miles/h
Time = 3.0 s [tex]=\dfrac{3}{3600}[/tex] hour
What is the acceleration of the car?
According to the first equation of motion, acceleration can be written as,
[tex]a=\dfrac{v-u}{t}[/tex]
substituting the values we get,
[tex]a=\dfrac{50-71}{\dfrac{3}{3600}}[/tex]
[tex]a=-25,200\rm\ miles/h^2[/tex]
Thus, the acceleration of the car is -25,200 miles/h².
What distance does the car travel during the braking period?
According to the third equation of motion,
[tex]v^2-u^2=2as[/tex]
Substituting the values we get,
[tex](50)^2-(71)^2=2(-25200)s[/tex]
[tex]s = 0.0504 \rm\ miles[/tex]
Thus, the distance car travel during the braking period is 0.0504 miles.
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