Answer to Question #207225 in Differential Geometry | Topology for Nestor Freeman

Question

The position vector r of a moving particle at time t after the start of the motion is given by

r =5(1+4t)i+5(19+2t−t2)j.

a) Find the initital velocity of the particle.

b) At time t = T, the particle is moving at right angles to its initial direction of motion. Find the value of T and the distance of the particle from its initial position at this time.

Answer 1

The position vector $r$ of a moving particle at time t after the start of the motion is given by

$r =5(1+4t)i+5(19+2t−t^2)j.$

a) It follows that the velocity of the particle is given by $v=r'_t =20i+5(2−2t)j.$ Therefore, the initital velocity of the particle is $v_0=20i+10j.$

b) Since at time $t = T,$ the particle is moving at right angles to its initial direction of motion, it folows that the dot product $r(0)\cdot r(T)=0,$ and hence $5\cdot5(1+4T)+95\cdot 5(19+2T−T^2)=0.$ Therefore, $1+4T+19(19+2T−T^2)=0,$ and thus $1+4T+361+38T−19T^2=0.$ It follows that $19T^2-42T-362=0,$ and $T>0$ implies $T=\frac{42+\sqrt{42^2+4\cdot 19\cdot 362}}{38}=\frac{42+\sqrt{29,276}}{38}\approx 5.6.$

The distance of the particle from its initial position to the position at time $T$ is equal

$5\sqrt{(4T)^2+(2T-T^2)^2} =5T\sqrt{16+(2-T)^2} =5T\sqrt{20-4T+T^2} \approx 150.7.$