5. In this question i and j are perpendicular horizontal unit vectors.] A particle P is moving with constant acceleration . At 2pm, the velocity of p is (3i+5j)kmh^-1 and at 2.30 pm the velocity of P is (i+7j)kmh^-1 At time T hours after 2 pm, P is moving in the direction of the vector (-i+2j) (a) Find the value of T. Another particle,Q.has velocity V_(Q) km h^-1 at time t hours after 2 pm , where v_(Q)=(-4-2t)i+(mu +3t)j and u is a constant. Given that there is an instant when the velocity of P is equal to the velocity of 9, (b) find the value of mu (6)

To find the value of T, we need to use the concept of velocity and acceleration. <br /><br />Given that the particle P is moving with constant acceleration, we can find the acceleration vector by taking the difference between the final and initial velocities and dividing it by the time taken.<br /><br />The initial velocity at 2 pm is $(3i+5j)kmh^{-1}$ and the final velocity at 2.30 pm is $(i+7j)kmh^{-1}$. <br /><br />The time taken is 0.5 hours (30 minutes).<br /><br />So, the acceleration vector is:<br /><br />$a = \frac{(i+7j) - (3i+5j)}{0.5}$<br /><br />Simplifying this, we get:<br /><br />$a = -2i + 2j$<br /><br />Now, we need to find the velocity of P at time T hours after 2 pm. We know that the velocity of P is in the direction of the vector $(-i+2j)$. <br /><br />Let's represent the velocity of P as $V_P = (x i + y j)$. <br /><br />Since the velocity is in the direction of $(-i+2j)$, we can write:<br /><br />$\frac{x}{-1} = \frac{y}{2}$<br /><br />Simplifying this, we get:<br /><br />$2x = -y$<br /><br />Now, we can use the equation of motion to find the velocity of P at time T:<br /><br />$V_P = V_0 + aT$<br /><br />where $V_0$ is the initial velocity at 2 pm, which is $(3i+5j)kmh^{-1}$.<br /><br />Substituting the values, we get:<br /><br />$(x i + y j) = (3i+5j) + (-2i + 2j)T$<br /><br />Simplifying this, we get:<br /><br />$(x i + y j) = (3i+5j) + (-2i + 2j)T$<br /><br />Now, we can solve for T by equating the coefficients of i and j on both sides of the equation:<br /><br />$x = 3 - 2T$<br /><br />$y = 5 + 2T$<br /><br />Using the relationship $2x = -y$, we can substitute the values of x and y:<br /><br />$2(3 - 2T) = -(5 + 2T)$<br /><br />Simplifying this, we get:<br /><br />$6 - 4T = -5 - 2T$<br /><br />$2T = 11$<br /><br />$T = \frac{11}{2}$<br /><br />Therefore, the value of T is $\frac{11}{2}$ hours.<br /><br />To find the value of $\mu$, we need to equate the velocity of P with the velocity of Q.<br /><br />The velocity of P is given by:<br /><br />$V_P = (x i + y j)$<br /><br />The velocity of Q is given by:<br /><br />$V_Q = (-4-2t)i + (\mu + 3t)j$<br /><br />Equating the velocities, we get:<br /><br />$(x i + y j) = (-4-2t)i + (\mu + 3t)j$<br /><br />Simplifying this, we get:<br /><br />$x = -4 - 2t$<br /><br />$y = \mu + 3t$<br /><br />Using the relationship $2x = -y$, we can substitute the values of x and y:<br /><br />$2(-4 - 2t) = -(\mu + 3t)$<br /><br />Simplifying this, we get:<br /><br />$-8 - 4t = -\mu - 3t$<br /><br />$\mu = -8 + t$<br /><br />Now, we need to find the value of t when the velocities are equal. We can do this by equating the magnitudes of the velocities:<br /><br />$\sqrt{x^2 + y^2} = \sqrt{(-4-2t)^2 + (\mu + 3t)^2}$<br /><br />Simplifying this, we get:<br /><br />$\sqrt{(3 - 2T)^2 + (5 + 2T)^2} = \sqrt{(-4-2t)^2 + (\mu + 3t)^2}$<br /><br />Solving this equation, we get:<br /><br />$T = \frac{11}{2}$<br /><br />Substituting this value of T into the equation for $\mu$, we get:<br /><br />$\mu = -8 + \frac{11}{2}$<br /><br />$\mu = \frac{-16 + 11}{2}$<br /><br />$\mu = \frac{-5}{2}$<br /><br />Therefore, the value of $\mu$ is $\frac{-5}{2}$.