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

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)

$a = \frac{\Delta v}{t}$

$a = \frac{v_f-v_i}{t}$

Position is given as,

$x = v_i t +\frac{1}{2} a t^2$

Replacing,

$x = v_i t +\frac{1}{2} (\frac{v_f-v_i}{t})t^2$

$x = \frac{v_i t}{2}+\frac{v_f t}{2}$

Reorganizing to find the initial speed

$v_i = \frac{2x}{t}-v_f$

Replacing we have,

$v_i = \frac{2(75m)}{7.1s}-15.9m/s$

$v_i = 5.23m/s$

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

PART B)

The acceleration of the Antelope would be

$a = \frac{v_f-v_i}{t}$

$a = \frac{15.9m/s-5.23m/s}{7.1s}$

$a = 1.503m/s^2$

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