An antelope moving with constant acceleration covers the distance between two points 75.0 m apart in 6.20 s. Its speed as it passes the second point is 15.0 m/s.

(a) What is its speed at the first point? (b) What is the antelope’s acceleration?

To solve this problem, we need to use the equations of motion for an object with constant acceleration. These equations are:

1. v = u + at
2. s = ut + (1/2)at^2
3. v^2 = u^2 + 2as

where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- t is the time taken
- s is the distance covered

Let's solve part (a) first:

We are given:
- Distance between two points, s = 75.0 m
- Time taken, t = 6.20 s
- Final velocity at the second point, v = 15.0 m/s

First, we can find the acceleration using equation (3):
15.0^2 = u^2 + 2a(75.0)

Simplifying:
225.0 = u^2 + 150a

Next, we can find the initial velocity using equation (2):
75.0 = u(6.20) + (1/2)a(6.20)^2

Simplifying:
75.0 = 6.20u + 19.24a

Now we have two equations with two unknowns (u and a). We can solve this system of equations simultaneously to find the values.

Let's solve part (b) now:

To find the acceleration, we can substitute the value of u from part (a) into either of the previous equations and solve for a.

Once we have found the value of a, we can substitute it back into the equation in part (a) to find the initial velocity u.

Let's solve these equations now.