Question

A runner starts from rest and in 3 s reaches a speed of 8 m/s. If we assume that the speed changed at a constant rate (constant net force), what was the average speed during this 3 s interval? average speed = Incorrect: Your answer is incorrect. m/s

Answer 1

Answer:

The average speed of the runner is 4 m/s.

Explanation:

Hi there!

The average speed (a.s) is calculated by dividing the traveled distance (d) over the time needed to travel that distance (t):

a.s = d / t

So, let´s find the distance traveled in those 3 s. For that, we can use the equation of position of an object moving in a straight line with constant acceleration:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

If we place the origin of the frame of reference at the point where the runner starts, then, x0 = 0. Since the runner starts from rest, v0 = 0. So, the equation gets reduced to this:

x = 1/2 · a · t²

We have the time (3 s), so let´s find the acceleration. For that, we can use the equation of velocity of an object moving in a straight line with constant acceleration:

v = v0 + a · t

Where "v" is the velocity at a time "t". Since v0 = 0, then:

v = a · t

At t = 3 s, v = 8 m/s

8 m/s = a · 3 s

8/3 m/s² = a

So let´s find the position of the runner at t = 3 s (In this case, the position of the runner will be equal to the traveled distance):

x = 1/2 · a · t²

x = 1/2 · 8/3 m/s² · (3 s)²

x = 12 m

Now, we can calcualte the average speed:

a.s = d/t

a.s = 12 m / 3 s

a.s = 4 m/s

The average speed of the runner is 4 m/s.

Answer 2

The average speed of the runner in the interval of 3s is $4\,m/s$.

Given that the runner starts from rest and reaches a speed of $8\,m/s$. ie;

• Initial velocity $u = 0 \,m/s$
• Final velocity $v= 8\, m/s$

The time taken for this change to occur is 3 seconds, ie;

• Time taken $t= 3s$

Therefore, we can find the acceleration,

• $a = \frac{v-u}{t} = \frac{8\,m/s \,- \,0\,m/s}{3 \,s} = \frac{8}{3} \, m/s^2$

We know that the average speed $(v_{av})$ is the total distance covered divided by the total time taken.

• ie; $v_{av} = \frac{s}{t}$

We can find the total distance using the kinematics equation;

• $s= ut+\frac{1}{2} at^2$

Substituting the known values, we get,

• $s = (0\times3)+(\frac{1}{2} \times \frac{8}{3} \,m/s^2 \times 3\,s\times 3\,s)= 12\,m$
• Now, the average speed $v_{av}=\frac{12\,m}{3\,s} = 4\,m/s$

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