maths mechanics

A particle P is moving with constant velocity. The position vector of P at time t seconds (t 0) is r metres, relative to a fixed origin O, and is given by
r = (2t – 3)i + (4 – 5t)j

(a) Find the initial position vector of P. =-3i+4j

The particle P passes through the point with position vector (3.4i – 12j)m at time T seconds.

(b) Find the value of T. =3.2 seconds
(c) Find the speed of P.

im not sure how to work out c
any help?
answer=v =2i + 5j

A few things not right there...2i + 5j is a vector, so it would have to be velocity, not speed as asked for in the question. Also if your formula for r is correct then the velocity would have to be 2i - 5j as the previous poster said

Oh I didn’t see it asked for “speed” -

The position vector r of the particle P can be thought of as a vector in a 2-dimensional real vector space, and the function r(t) that describes the motion of the particle as a linear transformation from the real line (which we can consider as a 1-dimensional real vector space) to this 2-dimensional space.

The position vector r of the particle P can be thought of as a vector in a 2-dimensional real vector space, and the function r(t) that describes the motion of the particle as a linear transformation from the real line (which we can consider as a 1-dimensional real vector space) to this 2-dimensional space.

(a) The initial position vector of P is found by applying the linear transformation r at time t = 0. By plugging t = 0 into the function, we obtain r(0) = -3i + 4j, which is indeed the initial position vector of P.

(b) Next, we want to find the time T at which the particle P passes through the point with position vector (3.4i - 12j). This is equivalent to solving the equation r(T) = 3.4i - 12j for T. By setting the components equal, we can derive two equations: 2T - 3 = 3.4 and 4 - 5T = -12. Solving this system of linear equations, we find T = 3.2 seconds.

(c) Lastly, we aim to find the speed of the particle P. In the language of differential geometry, the velocity vector of the particle is given by the derivative of the position vector with respect to time, v = dr/dt. The magnitude of the velocity vector is then the speed of the particle, which can be calculated by the Pythagorean theorem as sqrt[(2)^2 + (-5)^2] = sqrt[29] m/s.

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