At time t seconds, a particle P has velocity v m s⁻¹, where v = 3ti^ + 2tj^, t > 0 (a) Find the acceleration of P at time t seconds, where t > 0. (b) Find the value of t at the instant when P is moving in the direction of i - j. At time t seconds, where t > 0, the position vector of P, relative to a fixed origin O, is r metres. When t = 1, r = -j. (c) Find an expression for r in terms of t. (d) Find the exact distance of P from O at the instant when P is moving with speed 10 m s⁻¹.

A-Level Edexcel Maths Mechanics Variable Acceleration - 1D: At time t seconds, a particle P

At time t seconds, a particle P has velocity v m s⁻¹, where v = 3ti^ + 2tj^, t > 0 (a) Find the acceleration of P at time t seconds, where t > 0 - Edexcel - A-Level Maths Mechanics - Question 5 - 2021 - Paper 1

Question 5

At-time-t-seconds,-a-particle-P-has-velocity-v-m-s⁻¹,-where--v-=-3ti^-+-2tj^,--t->-0--(a)-Find-the-acceleration-of-P-at-time-t-seconds,-where-t->-0-Edexcel-A-Level Maths Mechanics-Question 5-2021-Paper 1.png

At time t seconds, a particle P has velocity v m s⁻¹, where v = 3ti^ + 2tj^, t > 0 (a) Find the acceleration of P at time t seconds, where t > 0. (b) Find the va... show full transcript

Worked Solution & Example Answer:At time t seconds, a particle P has velocity v m s⁻¹, where v = 3ti^ + 2tj^, t > 0 (a) Find the acceleration of P at time t seconds, where t > 0 - Edexcel - A-Level Maths Mechanics - Question 5 - 2021 - Paper 1

Step 1

Find the acceleration of P at time t seconds, where t > 0.

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Answer

To find the acceleration, we need to differentiate the velocity vector v with respect to time t:

rac{dv}{dt} = rac{d}{dt}(3ti^ + 2tj^)

This gives:

rac{dv}{dt} = 3i + 2j

Thus, the acceleration of particle P at time t seconds is:

Acceleration: 3i + 2j.

Step 2

Find the value of t at the instant when P is moving in the direction of i - j.

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Answer

For P to be moving in the direction of i - j, we normalize the vector (i - j):

ext{Unit vector} = rac{i - j}{ ext{Magnitude of }(i - j)} = rac{i - j}{\\sqrt{1^2 + (-1)^2}} = rac{i - j}{\\sqrt{2}}

The velocity vector must be a scalar multiple of this unit vector:

v = 3ti^ + 2tj^ = k rac{i - j}{\\sqrt{2}}, ext{ for some scalar } k.

Setting components equal gives:

3t = rac{k}{\\sqrt{2}} ext{ and } 2t = - rac{k}{\\ ext{Magnitude of }(i - j)}.

Solving these equations, we find:

3t + 2t = 0 ightarrow 5t = 0 ext{ so } , t = rac{9}{4}.

Step 3

Find an expression for r in terms of t.

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Answer

The position vector r can be found by integrating the velocity vector with respect to time:

r = rac{1}{2} (3t^2 i + 2t^2 j) + C,

where C is a constant vector. From the information given, when t = 1, r = -j:

r = -j = rac{1}{2} (3 imes 1^2i + 2 imes 1^2j) + C ightarrow C = - rac{3}{2} i - rac{1}{2} j.

Thus, the expression for r in terms of t is:

r = (1.5t^2 - 3) i + (t^2 - rac{1}{2}) j.$

Step 4

Find the exact distance of P from O at the instant when P is moving with speed 10 m s⁻¹.

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Answer

Thus the exact distance of P from O is $rac{10 ext{ m}}{\\sqrt{13}}$ .

At time t seconds, a particle P has velocity v m s⁻¹, where v = 3ti^ + 2tj^, t > 0 (a) Find the acceleration of P at time t seconds, where t > 0 - Edexcel - A-Level Maths Mechanics - Question 5 - 2021 - Paper 1

Worked Solution & Example Answer:At time t seconds, a particle P has velocity v m s⁻¹, where v = 3ti^ + 2tj^, t > 0 (a) Find the acceleration of P at time t seconds, where t > 0 - Edexcel - A-Level Maths Mechanics - Question 5 - 2021 - Paper 1

Find the acceleration of P at time t seconds, where t > 0.

Find the value of t at the instant when P is moving in the direction of i - j.

Find an expression for r in terms of t.

Find the exact distance of P from O at the instant when P is moving with speed 10 m s⁻¹.

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