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A car sits in an entrance ramp to a freeway, waiting for a break in the traffic. The driver sees a small gap between a van and an 18-wheel truck and accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 19.0 m/s when it reaches the end of the ramp, which has length 126 m .what is the acceleration of the car?
How much time does it take the car to travel the length of the ramp?
The traffic on the freeway is moving at a constant speed of 23.0 . What distance does the traffic travel while the car is moving the length of the ramp?

Respuesta :

Answer:

Part a)

[tex]a = 1.43 m/s^2[/tex]

Part b)

[tex]t = 13.3 s[/tex]

Part c)

[tex]d = 305 m[/tex]

Explanation:

Part a)

Car start from rest and reached to final speed of 19 m/s when it will cover a distance of 126 m

So we will have

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

now we will have

[tex]19^2 - 0 = 2(a)(126)[/tex]

[tex]a = 1.43 m/s^2[/tex]

Part b)

in order to find the time taken by the car we can use another kinematics equation

[tex]x = (\frac{v_f + v_i}{2}) t[/tex]

[tex]126 = (\frac{19 + 0}{2})t[/tex]

[tex]t = 13.3 s[/tex]

Part c)

If the traffic on free way is moving with speed 23 m/s

so we can say that traffic will move by distance

[tex]d = v t[/tex]

[tex]d = (23)(13.3)[/tex]

[tex]d = 305 m[/tex]

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