An antelope moving with constant acceleration covers the distance 70.0 m between two points in time 6.60 s. its speed as it passes the second point is 14.3 m/s. part a
what is its speed at the first point?
express your answer with the appropriate units. part b
what is the acceleration?
express your answer with the appropriate units.

A) The speed at the first point is 6.91 m/s.

B) The acceleration is 1.12 m/s²

Explanation:

The equations of velocity and position of the antelope will be as follows:

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) If we place the origin of the frame of reference at the first point, we can say that at t = 6.60 the position of the antelope is 70.0 m and its velocity is 14.3 m/s. In this way, we will have 2 equations with 2 unknowns, the initial velocity (the velocity at the first point) and the acceleration.

Let´s start finding the speed at the first point:

v = v0 + a · t       (solving for "a")

(v - v0)/t = a

Replacing a = (v - v0)/t  in the equation of the position:

x = x0 + v0 · t + 1/2 · (v - v0)/t · t²             (x0 = 0)

x = v0 · t  + 1/2 · v · t - 1/2 · v0 · t

x - 1/2 · v · t = 1/2 · v0 · t

2/t · (x - 1/2 · v · t) = v0

2/6.60 s · (70.0 m - 1/2 · 14.3 m/s · 6.60s) = v0

v0 = 6.91 m/s

The speed at the first point is 6.91 m/s.

B) Using the equation of velocity

a = (v - v0)/t

a = (14,3 m/s - 6.91 m/s) / 6.60 s

a = 1.12 m/s²

The acceleration is 1.12 m/s²