Question #207225

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.


1
Expert's answer
2022-01-10T14:05:55-0500

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

r=5(1+4t)i+5(19+2tt2)j.r =5(1+4t)i+5(19+2t−t^2)j.


a) It follows that the velocity of the particle is given by v=rt=20i+5(22t)j.v=r'_t =20i+5(2−2t)j. Therefore, the initital velocity of the particle is v0=20i+10j.v_0=20i+10j.


b) Since at time t=T,t = T, the particle is moving at right angles to its initial direction of motion, it folows that the dot product r(0)r(T)=0,r(0)\cdot r(T)=0, and hence 55(1+4T)+955(19+2TT2)=0.5\cdot5(1+4T)+95\cdot 5(19+2T−T^2)=0. Therefore, 1+4T+19(19+2TT2)=0,1+4T+19(19+2T−T^2)=0, and thus 1+4T+361+38T19T2=0.1+4T+361+38T−19T^2=0. It follows that 19T242T362=0,19T^2-42T-362=0, and T>0T>0 implies T=42+422+41936238=42+29,276385.6.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 TT is equal

5(4T)2+(2TT2)2=5T16+(2T)2=5T204T+T2150.7.5\sqrt{(4T)^2+(2T-T^2)^2} =5T\sqrt{16+(2-T)^2} =5T\sqrt{20-4T+T^2} \approx 150.7.



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS