Author Topic: vector calculus and kinematics questions (Read 1038 times) Tweet Share

chuhi · « **on:** August 24, 2011, 05:16:20 pm »

I have a SAC coming up...I am stuck on these questions

1. A body moves with a position given by $r(t)=1/(t+1) i +t j,t>or=0$
a)Find the displacement of the body during the time interval t=0 to t=10
Answer = $-10/11i + 10j$
b) find the average velocity of the body for t[0,10]
Answer= $-0.0909i +j$
c)Show that the body will eventually travel at a constant velocity parallel to the unit vector j

I have no idea about c...I don't get what it is asking..

2. Two trains pass one another, travelling in opposite directions on parallel tracks. When the fronts of the trains are in line they are travelling at 12m/s and 16m/s and accelerating at 0.5m/s^2 and 1m/s^2 respectively. The length of each train is 136 metres. How long does it take the rear of the two trains to pass?
Answer:8s

3. Two cars A and B, each moving with constant acceleration, are travelling in the same direction along the parallel lanes of a divided road. When A passes B, the speeds are 64 and 48 km/h respectively. Three minutes later B passes A, traveling at 96 km/h. Find
a) the distance travelled by A and B at this instant(since they first passed) and the speed of A.
Answer: a)3600m, 80km/h
b) the instant at which both are moving with the same speed, and the distance between them at this time
Answer: b)90s after A passed B,200m

Thanks so much for taking your time looking at this!

Mao · « **Reply #1 on:** August 24, 2011, 06:34:42 pm »

1c.

$r'(t) = -\frac{1}{(t+1)^2} i + j$

as $t\to \infty$ , $r'(t) \to j$ , that is the velocity becomes $j$ . This velocity is constant in the j direction.

2. We use $x=ut + \frac{1}{2}at^2$ . Define train #1 travelling in the +ve direction and accelerating in the +ve direction. thus Train #2 is travelling in the -ve direction and accelerating in the -ve direction.

Then, the equation for the front of the trains are $x_1 = 12t + 0.25t^2$ and $x_2 = -16t - 0.5 t^2$ . We seek the time when the front of the trains are separated by 272m. That is, $x_1 - x_2 = 272$ . Thus, solve $(12t + 0.25t^2)-(-16t - 0.5 t^2)=272$ .

3. let's work in units of km and hr. Then, the time elapsed is 3 min = 0.05 hrs.
a) car B: $x=\frac{v+u}{2}t=\frac{96+48}{2}*0.05 = 3.6km$
car A: $x=\frac{v+u}{2}t$ , thus $3.6=\frac{64+v}{2}*0.05$ , solve for v.

b) car A: acceleration is $a=\frac{v-u}{t}$ , thus $a=\frac{80-64}{0.05} = 320\, km/hr^2$
Velocity is thus $v=64+320t$ , displacement is $x = 64t + 160t^2$

car B: acceleration is $960\, km/hr^2$ , velocity is thus $v=48+960t$ , displacement is $x=48t + 480t^2$

Equate velocities to find they are equal when $t=0.025\, hr = 1.5\, min$ . Substitute into displacement, then find the difference.

chuhi · « **Reply #2 on:** August 25, 2011, 08:32:35 pm »

thanks a lot!:)

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