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6. At time t seconds, where t > 0, a particle P moves in the x-y plane in such a way that its velocity v m s⁻¹ is given by $$ extbf{v} = t extbf{i} - 4 extbf{j}$$ When t = 1, P is at the point A and when t = 4, P is at the point B - Edexcel - A-Level Maths Mechanics - Question 6 - 2018 - Paper 2

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6.-At-time-t-seconds,-where-t->-0,-a-particle-P-moves-in-the-x-y-plane-in-such-a-way-that-its-velocity-v-m-s⁻¹-is-given-by--$$-extbf{v}-=-t--extbf{i}---4-extbf{j}$$--When-t-=-1,-P-is-at-the-point-A-and-when-t-=-4,-P-is-at-the-point-B-Edexcel-A-Level Maths Mechanics-Question 6-2018-Paper 2.png

6. At time t seconds, where t > 0, a particle P moves in the x-y plane in such a way that its velocity v m s⁻¹ is given by $$ extbf{v} = t extbf{i} - 4 extbf{j}$$ ... show full transcript

Worked Solution & Example Answer:6. At time t seconds, where t > 0, a particle P moves in the x-y plane in such a way that its velocity v m s⁻¹ is given by $$ extbf{v} = t extbf{i} - 4 extbf{j}$$ When t = 1, P is at the point A and when t = 4, P is at the point B - Edexcel - A-Level Maths Mechanics - Question 6 - 2018 - Paper 2

Step 1

Find Position of Point A (t = 1)

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Answer

To find the position of point A at t = 1, we first need to calculate the position by integrating velocity with respect to time. The velocity extbfv=textbfi−4extbfj extbf{v} = t extbf{i} - 4 extbf{j} indicates

extbfr(t)=∫extbfvdt=∫(textbfi−4extbfj)dt=t22extbfi−4textbfj+extbfC extbf{r}(t) = \int extbf{v} dt = \int (t extbf{i} - 4 extbf{j}) dt = \frac{t^2}{2} extbf{i} - 4t extbf{j} + extbf{C}

At t = 1: extbfr(1)=(1)22extbfi−4(1)extbfj+extbfC=12extbfi−4extbfj+extbfC extbf{r}(1) = \frac{(1)^2}{2} extbf{i} - 4(1) extbf{j} + extbf{C} = \frac{1}{2} extbf{i} - 4 extbf{j} + extbf{C} Assuming point A is at the origin (0, 0), we have C=−12extbfi+4extbfj\textbf{C} = -\frac{1}{2} extbf{i} + 4 extbf{j}. Therefore, r(1)=0\textbf{r}(1) = 0 gives point A as ( A(0, 0) ).

Step 2

Find Position of Point B (t = 4)

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Answer

At t = 4: extbfr(4)=(4)22extbfi−4(4)extbfj+C=8extbfi−16extbfj+(−12extbfi+4extbfj) extbf{r}(4) = \frac{(4)^2}{2} extbf{i} - 4(4) extbf{j} + \textbf{C} = 8 extbf{i} - 16 extbf{j} + (-\frac{1}{2} extbf{i} + 4 extbf{j}) =(8−12)extbfi+(−16+4)extbfj=152extbfi−12extbfj= (8 - \frac{1}{2}) extbf{i} + (-16 + 4) extbf{j} = \frac{15}{2} extbf{i} - 12 extbf{j} Thus, point B is at ( B(7.5, -12) ).

Step 3

Calculate Distance AB

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Answer

The distance between points A and B is given by the distance formula: AB=(xB−xA)2+(yB−yA)2AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} Substituting the coordinates of A and B: AB=(7.5−0)2+(−12−0)2=(7.5)2+(−12)2=56.25+144=200.25=14.14AB = \sqrt{(7.5 - 0)^2 + (-12 - 0)^2} = \sqrt{(7.5)^2 + (-12)^2} = \sqrt{56.25 + 144} = \sqrt{200.25} = 14.14 Thus, the exact distance AB is ( AB = 14.14 ) units.

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