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The driver of a car wishes to pass a truck that is traveling at a constant speed of 20.5 m/s . initially, the car is also traveling at a speed 20.5 m/s and its front bumper is a distance 23.9 m behind the truck's rear bumper. the car begins accelerating at a constant acceleration 0.640 m/s2 , then pulls back into the truck's lane when the rear of the car is a distance 26.3 m ahead of the front of the truck. the car is of length 4.90 m and the truck is of length 20.2 m

Respuesta :

total distance moved by the car

[tex]d = 23.9 + 20.2 + 4.90 + 26.3 [/tex]

[tex]d = 75.3 m[/tex]

now given that the acceleration of the car is

a = 0.640 m/s^2

initial speed of both is given same so relative speed is zero

now we can use kinematics

[tex] d = \frac{1}{2} at^2[/tex]

[tex] 75.3 = \frac{1}{2}* 0.640 * t^2[/tex]

[tex] t = 15.33 s[/tex]

so it will take 15.33 s to overtake and reach the given distance

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