1.13 Speed  (Page 2/2)

Average speed

Problem : The object is moving with two different speeds $v_1$ and $v_2$ in two equal time intervals. Find the average speed.

Solution : The average speed is given by :

$$\begin{array}{c}{v}_a=\frac{\Delta s}{\Delta t}\end{array}$$

Let the duration of each time interval is t. Now, total distance is :

$$\begin{array}{c}\Delta s=v_{1}t+v_{2}t\end{array}$$

The total time is :

$$\begin{array}{c}\Delta t=t+t=\mathrm{2t}\end{array}$$

$$\begin{array}{ccc}\Rightarrow & v_a=& \frac{v_{1}t+v_{2}t}{\mathrm{2t}}\end{array}$$

$$\begin{array}{ccc}\Rightarrow & v_a=& \frac{v_1+v_2}{2}\end{array}$$

The average speed is equal to the arithmetic mean of two speeds.

Problem The position – time plot of a particle’s motion is shown below.

Determine :

• average speed in the first 10 seconds
• Instantaneous speed at 5 seconds
• Maximum speed speed
• The time instant(s), when average speed measured from the beginning of motion equals instantaneous speed

Solution

1. Average speed in the first 10 seconds

The particle covers a distance of 110 m. Thus, average speed in the first 10 seconds is :

$$\begin{array}{l}{v}_a=\frac{110}{10}=\mathrm{11\; m/s}\end{array}$$

2. Instantaneous speed at 5 second

We draw the tangent at t = 5 seconds as shown in the figure. The tangent coincides the curve between t = 5 s to 7 s. Now,

$$\begin{array}{l}v=\mathrm{slope\; of\; the\; tangent}=\frac{30}{2}=\mathrm{15\; m/s}\end{array}$$

3. Maximum speed

We observe that the slope of the tangent to the curve first increase to become constant between A and B. The slope of the tangent after point B decreases to become almost flat at the end of the motion. It means the maximum speed corresponds to the constant slope between A and B, which is is 15 m/s

4. Time instant(s) when average speed equals instantaneous speed

Average speed is equal to slope of chord between one point to another. On the other hand, speed is equal to slope of tangent at point on the plot. This means that slope of chord between two times is equal to tangent at the end point of motion. Now, as time is measured from the start of motion i.e. from origin, we draw a straight line from the origin, which is tangent to the curve. The only such tangent is shown in the figure below. Now, this straight line is the chord between origin and a point. This line is also tangent to the curve at that point. Thus, average speed equals instantaneous speeds at t = 9 s.

Problem : Two particles are moving with the same constant speed, but in opposite direction. Under what circumstance will the separation between two remains constant?

Solution : The condition of motion as stated in the question is possible, if particles are at diametrically opposite positions on a circular path. Two particles are always separated by the diameter of the circular path. See the figure below to evaluate the motion and separation between the particles.