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A particle P is moving with constant velocity (-3i + 2j) m s^(-1) - Edexcel - A-Level Maths: Mechanics - Question 1 - 2010 - Paper 1

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A particle P is moving with constant velocity (-3i + 2j) m s^(-1). At time t = 6 s, P is at the point with position vector (-4i - 7j) m. Find the distance of P from ... show full transcript

Worked Solution & Example Answer:A particle P is moving with constant velocity (-3i + 2j) m s^(-1) - Edexcel - A-Level Maths: Mechanics - Question 1 - 2010 - Paper 1

Step 1

Find the position vector at time t = 2 s

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Answer

Since the particle moves with constant velocity, we can determine its position vector at any time t using the formula:

r(t)=r(6)+v(t−6)r(t) = r(6) + v(t - 6)

Where:

  • r(6)=−4i−7jr(6) = -4i - 7j m (position vector at t = 6 s)
  • v=−3i+2jv = -3i + 2j m/s (velocity)
  • t=2t = 2 s, hence (t−6)=−4(t-6) = -4 s.

So, substituting the values:

r(2)=(−4i−7j)+(−3i+2j)(2−6)r(2) = (-4i - 7j) + (-3i + 2j)(2 - 6) =(−4i−7j)+(−3i+2j)(−4)= (-4i - 7j) + (-3i + 2j)(-4) =(−4i−7j)+(12i−8j)= (-4i - 7j) + (12i - 8j) =(8i−15j)extm= (8i - 15j) ext{ m}

Step 2

Find the distance from the origin

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Answer

The distance from the origin to the position vector can be calculated using the magnitude of the position vector:

∣r∣=extsqrt((8)2+(−15)2)|r| = ext{sqrt}((8)^2 + (-15)^2)

Calculating this gives:

∣r∣=extsqrt(64+225)=extsqrt(289)=17extm|r| = ext{sqrt}(64 + 225) = ext{sqrt}(289) = 17 ext{ m}

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