# Thread: vector: motion in space

1. ## vector: motion in space

pls guys help me with this question

a beetle is climbing on a car. in its starting position,the surface of the car is represented by the unit semicircle x^2 + y^2 = 1, y>0 in the xy-plane. the road is represented as the x-axis. at time t=0 the beetle starts at the front bumper, (1,0) and crawls counterclockwise around the car at unit speed relative to the car. at the same time the car moves to the right at speed 10. find the parametric formula for the trajectory of the beetle,and find its position when it reaches the rear bumper. (at t=0 the rear bumper is at (-1,0).)

2. Originally Posted by jijosantos
pls guys help me with this question

a beetle is climbing on a car. in its starting position,the surface of the car is represented by the unit semicircle x^2 + y^2 = 1, y>0 in the xy-plane. the road is represented as the x-axis. at time t=0 the beetle starts at the front bumper, (1,0) and crawls counterclockwise around the car at unit speed relative to the car. at the same time the car moves to the right at speed 10. find the parametric formula for the trajectory of the beetle,and find its position when it reaches the rear bumper. (at t=0 the rear bumper is at (-1,0).)
the centre of the circle is at (10t,0) at time t.
the beetle was crawling at a speed of 1 unit w.r.t. Let the centre of the circle be called C. Let the beetle be at point B at time t. Then the anle which CB makes with the x-axis is t radians. (do you see why?)
the position of the beetle is hence (10t+cos(t), sin(t)). eleminate t and you will get the trajectory.

EDIT: for the position of the beetle when it reaches the rear bumper use the equation (10t + cos(t),sin(t))=(10t-1,0) which gives t= pi
so the position of the beetle is (10pi-1,0)

3. Originally Posted by abhishekkgp
the centre of the circle is at (10t,0) at time t.
the beetle was crawling at a speed of 1 unit w.r.t. Let the centre of the circle be called C. Let the beetle be at point B at time t. Then the anle which CB makes with the x-axis is t radians. (do you see why?)
the position of the beetle is hence (10t+cos(t), sin(t)). eleminate t and you will get the trajectory.

EDIT: for the position of the beetle when it reaches the rear bumper use the equation (10t + cos(t),sin(t))=(10t-1,0) which gives t= pi
so the position of the beetle is (10pi-1,0)
ok but i think the speed of the beetle is not constant at all. the tangent vector for the semicircle is not the same along the curve..i still dont get it where is the largest and smallest value of speed..is it depend on the y-axis value?

4. Originally Posted by jijosantos
ok but i think the speed of the beetle is not constant at all. the tangent vector for the semicircle is not the same along the curve..i still dont get it where is the largest and smallest value of speed..is it depend on the y-axis value?
you are right... the speed of the beetle is not constant with time.
to find the time at which the speed is max and min you need to find the speed as a function of time first.
to do this let $\displaystyle s(t)$ denote speed of the beetle at time $\displaystyle t$.
then [s(t)]^2= (dx/dt)^2 + (dy/dt)^2
this will give you speed as a function of time. differentiate the expression of speed w.r.t time and set it equal to zero. you will get the times at which the speed reaches maximum and minimum.

"please 'thank' me by clicking at the bottom left of my posts if you find this useful."

5. Originally Posted by abhishekkgp
you are right... the speed of the beetle is not constant with time.
to find the time at which the speed is max and min you need to find the speed as a function of time first.
to do this let $\displaystyle s(t)$ denote speed of the beetle at time $\displaystyle t$.
then [s(t)]^2= (dx/dt)^2 + (dy/dt)^2
this will give you speed as a function of time. differentiate the expression of speed w.r.t time and set it equal to zero. you will get the times at which the speed reaches maximum and minimum.

"please 'thank' me by clicking at the bottom left of my posts if you find this useful."
ok now i get it..thanks for your help