What Minimum Deceleration Ensures Legal Speed Over 116 Meters?

[shawonna23]

A speed trap is set up with two pressure-activated strips placed across a highway, 116 m apart. A car is speeding along at 31.5 m/s, while the speed limit is 21 m/s. At the instant the car activates the first strip, the driver begins slowing down. What minimum deceleration is needed in order that the average speed is within the limit by the time the car crosses the second marker?

I have no clue on what equation to use to solve this problem or how I would go about solving this problem. Can someone please help me?

 

[Nylex]

Look at the quantities you have and the one you're looking for and decide on the equation to use from those. It's one of the UVATS ones..

 

[wais]

To solve this problem, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this case, u = 31.5 m/s, v = 21 m/s, and t = t2 - t1, where t1 is the time when the car activates the first strip and t2 is the time when the car crosses the second strip.

We also know that the distance between the two strips is 116 m. Using the equation d = ut + 1/2at^2, we can find the time it takes for the car to travel between the two strips. Since the initial velocity is 31.5 m/s and the final velocity is 21 m/s, the average velocity is (31.5 + 21)/2 = 26.25 m/s. Substituting these values into the equation, we get:

116 = (31.5)(t2 - t1) + 1/2a(t2 - t1)^2

Simplifying this equation, we get:

116 = 31.5t2 - 31.5t1 + 1/2at2^2 - at1t2 + 1/2a(t1)^2

Since we know that t2 - t1 = t, we can substitute this into the equation to get:

116 = 31.5t + 1/2at^2 - at^2 + 1/2a(t)^2

Simplifying further, we get:

116 = 31.5t + 1/2at^2 - 1/2at^2 + 1/2a(t)^2

116 = 31.5t + 1/2a(t)^2

Now, we need to solve for the minimum deceleration (a) needed for the average speed to be within the limit. To do this, we need to set the average speed equal to the speed limit (21 m/s) and solve for a. The equation becomes:

21 = 31.5t + 1/2a(t)^2

Rearranging this equation, we get:

1/2a(t)^2 + 31.5t - 21 = 0

Using the quadratic formula, we can solve for t:

t = (-