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At time t = 0, a football player kicks a ball from the point A with position vector (2i + j) m on a horizontal football field - Edexcel - A-Level Maths Mechanics - Question 8 - 2005 - Paper 1

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At time t = 0, a football player kicks a ball from the point A with position vector (2i + j) m on a horizontal football field. The motion of the ball is modelled as ... show full transcript

Worked Solution & Example Answer:At time t = 0, a football player kicks a ball from the point A with position vector (2i + j) m on a horizontal football field - Edexcel - A-Level Maths Mechanics - Question 8 - 2005 - Paper 1

Step 1

a) the speed of the ball

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Answer

To find the speed of the ball, we use the formula for speed, which is the magnitude of the velocity vector. The given velocity vector is (5i + 8j) m/s.

Calculate the magnitude:

extSpeed=extMagnitude(5i+8j)=sqrt52+82=sqrt25+64=sqrt89approx9.43 extm/s ext{Speed} = ext{Magnitude}(5i + 8j) = \\sqrt{5^2 + 8^2}\\ = \\sqrt{25 + 64} = \\sqrt{89} \\approx 9.43 \, ext{m/s}

Thus, the speed of the ball is approximately 9.43 m/s.

Step 2

b) the position vector of the ball after t seconds

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Answer

The position vector of the ball can be determined using the initial position vector and adding the displacement due to velocity over time:

Given:

  • Initial position vector: (2i + j) m
  • Velocity: (5i + 8j) m/s

The displacement after t seconds is given by:

extDisplacement=extVelocity×t=(5i+8j) t ext{Displacement} = ext{Velocity} \times t = (5i + 8j) \, t

Thus, the position vector after t seconds is:

extPositionvector=(2i+j)+(5ti+8tj)=(2+5t)i+(1+8t)j ext{Position vector} = (2i + j) + (5ti + 8tj) = (2 + 5t)i + (1 + 8t)j

Step 3

c) Find the time when the ball is due north of B

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Answer

The point B has a position vector of (10i + 7j). For the ball to be due north of B, the x-components must be the same:

Let the position of the ball be (2 + 5t)i + (1 + 8t)j. Set the x-component equal to that of B:

2+5t=102 + 5t = 10

Solving for t gives:

5t=10−25t=8t=85=1.6 seconds5t = 10 - 2 \\ 5t = 8 \\ t = \frac{8}{5} = 1.6 \text{ seconds}

Step 4

d) find the value of v

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Answer

When the ball is at the position vector at t = 1.6 s, we substitute t into the expression for the position vector:

extPositionvector=(2+5(1.6))i+(1+8(1.6))j=(2+8)i+(1+12.8)j=10i+13.8j ext{Position vector} = (2 + 5(1.6))i + (1 + 8(1.6))j \\ = (2 + 8)i + (1 + 12.8)j \\ = 10i + 13.8j

From the point B (10i + 7j), we know that the second player starts running due north from B with speed v. The distance travelled can be calculated as follows:

The distance travelled when t = 1.6 seconds:

extDistance=13.8−7=6.8 extm ext{Distance} = 13.8 - 7 = 6.8 \, ext{m}

Thus, with speed v, distance travelled can also be represented as:

extDistance=v×t6.8=vimes1.6 ext{Distance} = v \times t \\ 6.8 = v imes 1.6

Solving for v:

v=6.81.6=4.25 extm/sv = \frac{6.8}{1.6} = 4.25 \, ext{m/s}

Step 5

e) State one physical factor, other than air resistance

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Answer

One physical factor that would be important for a more realistic model of the ball's motion is friction on the field. The effect of friction can alter the ball's velocity and trajectory, impacting how far and in what direction the ball travels over time.

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