Question

An antelope moving with constant acceleration covers the distance 75.0 m between two points in time 7.10 s. Its speed as it passes the second point is 15.9 m/s.

a. What is its speed at the first point?


b. What is the acceleration?

Answer 1

The initial speed and acceleration of the antelope can be found by using two kinematic equations that involve the final velocity, total distance covered, and the time taken. Solving these equations simultaneously yields the initial speed and acceleration.

Step-by-step explanation:

The student has provided the distance an antelope has covered, the time it took, and its final speed. To determine the initial speed of the antelope at the first point, we use the kinematic equation:

v_f = v_i + a*t

where v_f is the final velocity (15.9 m/s), v_i is the initial velocity, a is acceleration, and t is the time (7.10 s).

For the acceleration, we use the following equation:

d = v_i*t + (1/2)*a*t^2

where d is the distance (75.0 m).

We have two equations with two unknowns (v_i and a), which can be solved simultaneously. Rearranging the first equation to solve for a gives us:

a = (v_f - v_i) / t

Substituting the expression for a into the second equation allows us to find v_i and then a.

Answer 2

To solve this problem we will apply the concepts related to the kinematic equations of linear emotion. We will start from the definition of acceleration as the change of speed as a function of time. Later we will apply the second law of kinematics for which displacement is understood as a function of initial speed, time and acceleration. We will use the expressions given initially for acceleration as a function of velocity and we will obtain a system of equations that will allow us to find the initial velocity, and subsequently the acceleration

PART A)

`a = (\Delta v)/(t)`

`a = (v_f-v_i)/(t)`

Position is given as,

`x = v_i t +(1)/(2) a t^2`

Replacing,

`x = v_i t +(1)/(2) ((v_f-v_i)/(t))t^2`

`x = (v_i t)/(2)+(v_f t)/(2)`

Reorganizing to find the initial speed

`v_i = (2x)/(t)-v_f`

Replacing we have,

`v_i = (2(75m))/(7.1s)-15.9m/s`

`v_i = 5.23m/s`

Therefore the speed of Antelope at first point is 5.23m/s

PART B)

The acceleration of the Antelope would be

`a = (v_f-v_i)/(t)`

`a = (15.9m/s-5.23m/s)/(7.1s)`

`a = 1.503m/s^2`

Therefore the acceleration of the Antelope is
`1.503m/s^2`