Question

MEDAL GIVEN TO CORRECT ANSWER.

A soccer player runs across a soccer field.

a. If the soccer player runs with a speed of 4.6 m/s, how long does it take him to run 60 m?

b. It takes the soccer player 3 seconds to reduce his speed to 3.1 m/s. What is his acceleration?

c. The soccer player is running at a speed of 4 m/s when he slides on the ground, coming to a stop in a distance of 2 m. What was his acceleration?

d. The soccer player accelerates from rest at a rate of 3.1 m/s2. How far does he run in 1.5 seconds?

Answer 1

Here are the equations you need for each question:
a)
$$
s=vt\\
\implies t=\cfrac{s}{t}
$$
b)
$$
a=\cfrac{v_f-v_i}{\Delta t}
$$
c)
$$
s=\cfrac{v_i+v_f}{2}\Delta t\\
=\cfrac{v_i+0}{2}\Delta t\\
\implies \Delta t=\cfrac{s}{\cfrac{v_i}{2}}\\
a=\cfrac{v_f-v_i}{\Delta t}\\
a=\cfrac{0-v_i}{\Delta t}
$$
d)
$$
s=\cfrac{v_i+v_f}{2}\Delta t\\
=\cfrac{0+v_f}{2}\Delta t
$$

Now just fill in the variables with your givens.

Make sense?

Answer 2

you just confused me even more, tbh.

Answer 3

Time is distance divided by speed.

Acceleration is change in velocity divided by time.

Stopping a velocity v requires a negative acceleration such that
v squared = twice the distance times the acceleration


distance from stopped position is (1/2)(acceleration)(ime)^2

Answer 4

Plugging in values, we get:

Question a
$$
s=vt\\
\implies t=\cfrac{s}{v}\\
t=\cfrac{60}{4.6} \text{s}
$$
Question b
$$
a=\cfrac{v_f-v_i}{\Delta t}\\
=\cfrac{3.1-4.6}{3}\cfrac{\text{m }}{\text{s}^2}
$$
Question c
$$
s=\cfrac{v_i+v_f}{2}\Delta t\\
=\cfrac{v_i+0}{2}\Delta t\\
\implies \Delta t=\cfrac{s}{\cfrac{v_i}{2}}\\
\Delta t=\cfrac{2}{\cfrac{3}{2}}=\cfrac{1}{3}\text{ s}\\
a=\cfrac{v_f-v_i}{\Delta t}\\
a=\cfrac{0-v_i}{\Delta t}\\
a=\cfrac{0-4}{\cfrac{1}{3}}=-12\cfrac{\text{m }}{\text{s}^2}
$$
Question d
$$
s=\cfrac{v_i+v_f}{2}\Delta t\\
=\cfrac{0+v_f}{2}\Delta t\\
=\cfrac{3.1}{2}1.5\text{ m}\\
$$