An object moving in a certain medium, with speed vv m/s, experiences a resistance to motion of RR newtons. RR and vv are related by the equation R=av2+bvR = av^2 + bv, where aa and bb are constants. The table shows the experimental values of the variables vv and RR.
vv510152025
RR174481128185

  1. (i)

    On the grid on page 11, plot Rv\frac{R}{v} against vv and draw a straight line graph.

  2. (ii)

    Use your graph to estimate the value of aa and of bb.

  3. (iii)

    In a different medium, RR is directly proportional to vv and R=30R = 30 when v=5v = 5.

    Draw a suitable line on your graph to illustrate the second situation and use it to determine the value of vv for which the resistance is the same in both mediums.

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Worked solution

5(i)

1. Calculate the values of $\frac{R}{v}$ for each corresponding $v$.

For v=5v = 5, Rv=175=3.4\frac{R}{v} = \frac{17}{5} = 3.4
For v=10v = 10, Rv=4410=4.4\frac{R}{v} = \frac{44}{10} = 4.4
For v=15v = 15, Rv=8115=5.4\frac{R}{v} = \frac{81}{15} = 5.4
For v=20v = 20, Rv=12820=6.4\frac{R}{v} = \frac{128}{20} = 6.4
For v=25v = 25, Rv=18525=7.4\frac{R}{v} = \frac{185}{25} = 7.4

vv510152025
Rv\frac{R}{v}3.44.45.46.47.4

2. Plot the points on the grid and draw the straight line.

Choose suitable scales for the vv-axis (horizontal) and Rv\frac{R}{v}-axis (vertical).
Plot the points (5,3.4)(5, 3.4), (10,4.4)(10, 4.4), (15,5.4)(15, 5.4), (20,6.4)(20, 6.4), and (25,7.4)(25, 7.4).
Draw a straight line passing through all these points.

Answer

A straight line graph plotted with points (5,3.4)(5, 3.4), (10,4.4)(10, 4.4), (15,5.4)(15, 5.4), (20,6.4)(20, 6.4), and (25,7.4)(25, 7.4).

5(ii)

1. Express the given equation in linear form

Given R=av2+bvR = av^2 + bv, dividing by vv gives:
Rv=av+b\frac{R}{v} = av + b
This is in the form Y=mX+cY = mX + c, where Y=RvY = \frac{R}{v}, X=vX = v, gradient m=am = a, and vertical intercept c=bc = b.

2. Calculate the gradient to find $a$

Using two points from the line, e.g., (5,3.4)(5, 3.4) and (25,7.4)(25, 7.4):
a=gradient=7.4−3.425−5=420=0.2a = \text{gradient} = \frac{7.4 - 3.4}{25 - 5} = \frac{4}{20} = 0.2

3. Find the vertical intercept to determine $b$

From the graph, the vertical intercept (where v=0v = 0) is 2.42.4.
Alternatively, using a point (5,3.4)(5, 3.4):
3.4=0.2(5)+b  ⟹  b=3.4−1=2.43.4 = 0.2(5) + b \implies b = 3.4 - 1 = 2.4

Answer

a=0.2a = 0.2, b=2.4b = 2.4

5(iii)

1. Find the equation relating $R$ and $v$ for the second medium.

Since RR is directly proportional to vv, R=kvR = kv.
Given R=30R = 30 when v=5v = 5:
30=k(5)  ⟹  k=630 = k(5) \implies k = 6
So, R=6vR = 6v.

2. Determine the equation of the line to be drawn on the $\frac{R}{v}$ against $v$ graph.

Divide the equation by vv to match the yy-axis of the graph:
Rv=6\frac{R}{v} = 6
This is a horizontal line passing through 66 on the vertical axis.

3. Find the value of $v$ where the resistance is the same.

The resistance is the same when the two graphs intersect.
From the graph of Rv\frac{R}{v} against vv, the line Rv=6\frac{R}{v} = 6 intersects the original line at v=18v = 18.

Answer

Draw the horizontal line Rv=6\frac{R}{v} = 6. The value is v=18v = 18.

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