Question

1. A subway train is travelling at 80km/h when it approaches a slower train 50 m ahead
travelling in the same direction at 25 km/h. if the faster train begins to decelerate at 2.6
m/s2
. While the slower train continues at constant speed. How soon and what relative
speed will they collide. [8]

2. A farm decided to construct a portion of his area close to his house for chicken farming.
He decided to fence a rectangular area of length 4x and width of 3/2x^2, the length is
measured with uncertainty of 12 cm and width of 24 cm uncertainty. If the maximum
perimeter of this area is 350 m.
a) Find the length and width [6]
b) Calculate the area of this portion of land [2]
c) Calculate the percentage of uncertainty of the minimum area
[3]

Answer 1

1) Let $V_1$ be the velocity of the first train and $V_2$ be the velocity of the second train. We know that $V_1(0)=80\,km/h =80000\,m/h$

and

$V_2=25\,km/h =25000\,m/h.$

Let $\Delta V(t) = V_1(t)-V_2(t)$ be the relative velocity of trains. If $a$ is deceleration (so it is a negative value), we get

$V_1(t) = V_1(0) +at.$ (1)

Therefore, the relative velocity can be written as

$\Delta V(t) = V_1(0) +at-V_2 = (V_1(0)-V_2) + at.$

We know the initial distance and we should calculate the time t, at which the distance will be equal to 0. For a motion with constant acceleration or deceleration with initial velocity $(V_1(0)-V_2)$ we get

$S=(V_1(0)-V_2)T + \dfrac{aT^2}{2},$ (2)

where T is the time from the beginning till the collision.

We may rewrite this formula using (1) and get

$S = \dfrac{(\Delta V(T))^2-(V_1(0)-V_2)^2}{2a},$

where $\Delta V(T)$ is the relative velocity of 1st train in a moment of collision.

Therefore

$(\Delta V(T))^2 =2aS+(V_1(0)-V_2)^2$ and $\Delta V(T) =\sqrt{2aS+(V_1(0)-V_2)^2} = \sqrt{-2\cdot 2.6\,m/s^2\cdot50\,m + \Big(\dfrac{80000\,m/h-25000\,m/h}{3600\,s}\Big)^2} = \sqrt{-26/6\,m^2/s^2}.$

This result is not confusing, because the modulus of deceleration is too big, so trains could nor collide.

If we rewrite the second equation as

$\dfrac{aT^2}{2} + (V_1(0)-V_2)T -S = 0$ ,

and calculate the discriminant

$D = (V_1(0)-V_2)^2 +4\dfrac{a}{2}S = \Big(\dfrac{80000\,m/h-25000\,m/h}{3600\,s}\Big)^2 - 2\cdot2.6\,m/s^2\cdot50\,m = -26.6\,m^2/s^2,$

we see that the equation has no real roots! So the answer is:the trains never collide for such a big deceleration.

2) If $l$ is length of the area and $w$ is the width of the area, then the perimeter is

$p = 2(l+w).$

Then we get an equation

$p = 2\Big(4x + \dfrac{3x^2}{2}\Big) = 8x + 3x^2.$

a) we know that $p=350\,m.$ So we should find the non-negative roots of an equation

$3x^2+8x-350 = 0$ ,

we get $x_1 = 9.55, \quad x_2 = -12.22.$

So we choose $x=9.55.$ Therefore $l=4x=4\cdot9.55=38.2\,m , \,\, w = \dfrac{3x^2}{2} = \dfrac{3\cdot9.55^2}{2} =136.8\,m.$

Note that if we decide that the value of 350 accounts the uncertainty of the measurements, we will get an equation

$p = 2\Big(4x + 0.12 + \dfrac{3x^2}{2} + 0.24 \Big) = 8x + 3x^2 + 0.72.$

Solving this equation, we get

$x_1 = 9.54, \quad x_2 = -12.21,$

so $l=4\cdot9.54 = 38.16\,m, \;\; w = \dfrac{3\cdot9.54^2}{2}=136.5\,m.$

b) The area can be calculated as $S=l\cdot w = 38.2\,m\cdot136.8\,m = 5225.8\,m^2.$ If we take the uncertainty

c) Let us determine the percentage uncertainty in the length and width (see also https://qualifications.pearson.com/content/dam/pdf/A%20Level/Physics/2015/Specification%20and%20sample%20assessments/Appendix%2010%20Updated.pdf page 4).

Percentage uncertainty in length $= \dfrac{0.12\,m}{38.2\,m} = 0.00314 \approx 0.31\%,$ percentage uncertainty in width $=\dfrac{0.24\,m}{136.8\,m} = 0.00175 \approx 0.18\%.$

If we get the percentage uncertainty of the product of two values, we should add the percentage uncertainties in length and in width to get $0.31\%+0.18\% = 0.49\%.$