An antelope moving with constant acceleration covers the distance 68.0 m between two points in time 7.50 s. its speed as it passes the second point is 15.7 m/s. a. what is its speed at the first point?
express your answer with the appropriate units.
b. what is the acceleration?
express your answer with the appropriate units.

A)The speed of the antelope at the first point is 2.43 m/s.

B) The acceleration of the antelope is 1.77 m/s²

Explanation:

The equations of the position and velocity of the antelope is given by the following expressions:

x = x0 + v0 · t + 1/2 · a ·t²

v = v0 + a · t

Where:

x = position of the antelope at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

A) Let´s place the center of the frame of reference at the first point. The equation of position at t = 7.50 s will be:

x = x0 + v0 · t + 1/2 · a ·t²

68.0 m = 0 m + v0 · 7.50 s + 1/2 · a · (7.50 s)²

We also know that at the second point the velocity is 15.7 m/s. Then at t = 7.50 the velocity will be 15.7 m/s.

v = v0 + a · t

15.7 m/s = v0 + a · 7.50 s

We can solve this equation for "a" and replace it in the equation of height to obtain "v0". Then:

a = (15.7 m/s - v0) / 7.50 s

Replacing it in the equation for position:

68.0 m = 0 m + v0 · 7.50 s + 1/2 · a · (7.50 s)²

68.0 m = v0 · 7.50 s + 1/2 · (15.7 m/s - v0) / 7.50 s · (7.50 s)²

68.0 m = v0 · 7.50 s + 7.85 m/s · 7.50 s - 3.75 s · v0

68.0 m - 7.85 m/s · 7.50 s = 3.75 s · v0

(68.0 m - 7.85 m/s · 7.50 s) / 3.75 s = v0

v0 = 2.43 m/s

The speed of the antelope at the first point is 2.43 m/s.

B) The acceleration of the antelope will be:

a = (15.7 m/s - v0) / 7.50 s

a = (15.7 m/s - 2.43 m/s) / 7.50 s

a = 1.77 m/s²

The acceleration of the antelope is 1.77 m/s²