How wide can a 5 m load be to fit under a parabolic underpass that is 7 m wide and 7 m at its highest point

Step-by-step explanation:

This is equivalent to finding the equation of a parabola from three points.

Let's call the ground-level points (-3.5, 0) and (3.5, 0).

The axis of symmetry will be the y-axis, so the vertex is at (0, 7).

Part 1. Find the equation of the parabola

The vertex form of the equation for a parabola is

y = a(x - h)² + k

where (h, k) is the vertex of the parabola.

If the vertex is at (0, 7),  

h = 0 and k = 7

The equation is

y = ax² + 7

Insert the point (0, -3.5)

0 = a(0 + 3.5)² + 7

0 = a(3.5)² + 7

0 = 12.25a + 7

-7 = 12.25a

a = -0.5714

The equation in vertex form is

y = -0.5714x² + 7

Part 2. Calculate the width when y = 5

 5 = -0.5714x² + 7

-2 = -0.5714x²

x² = 2/0.5714

x² = 3.5

x = ±√3.5 = ±1.87  

So, on the graph, the coordinates of the truck at ground level are x = -1.87 and x = +1.87.

The maximum width of the load is 1.87 – (-1.87) =

In the diagram below, the red parabola represents the underpass, and the blue box is the truck going through it.

i think it will be zero as slope has a formula of difference between initial and final distance divided by itz time interval