Two dimensional motion with variable acceleration

calculusclassical-mechanicsintegration

I'm doing a question in a textbook which is known to sometimes have wrong answers. However, it's more likely that I'm just being stupid.

Particles P and Q, each of mass 0.5 kg, move on a horizontal plane, with east and north as i and j directions. (x and y).

Initially, P has velocity (2i -5j)m/s and Q is travelling north at 2m/s. Each particle is acted on by a force of magnitude t Newtons.

The force on P acts towards north-east, while that on Q acts towards the south-east. Work out the value of t for which:

a. The two particles have the same speed.

b. The two particles are travelling in the same direction.

First of all I'm confused about t. I'm sure t in the question is the force, and can be split into components of the force. However in the mark scheme I'm told to integrate with respect to t, which implies t means time.

For part a I integrated the acceleration with respect to t anyway and got the velocity at a time. Then I understand that you can find the constants for both velocity of P and Q and then equate them to find the time when they both have the same speed. I managed to get to the answer in the textbook which is 4.2 seconds.

Now part B is where I'm completely stuck and the mark scheme has no explanation of what they are doing to get the answer which is apparently 2.11 seconds.

How do you show that two particles are travelling in the same direction and how can I find the time when this occurs?

Is the criteria such that the velocities need to be in the same direction? Any help would be appreciated.

Best Answer

$$F(t)=0.5a(t)$$

$$a(t)=2F(t)$$ That is $$a_P(t)=\sqrt2t(i+j)$$

and $$a_Q(t)=\sqrt2t(i-j)$$

Integrating the acceleration, $$v_P(t)=v_P(0)+\int_0^t a_P(s)\, ds=(2i-5j)+\frac{t^2}{\sqrt2}(i+j)$$

$$v_Q(t)=v_Q(0)+\int_0^t a_Q(s)\, ds=(2j)+\frac{t^2}{\sqrt2}(i-j)$$

Hence, we are interested in when does $$\frac{2+\frac{t^2}{\sqrt2}}{-5+\frac{t^2}{\sqrt2}}=\frac{\frac{t^2}{\sqrt2}}{2-\frac{t^2}{\sqrt2}}$$

$$4-\frac{t^4}{2}=-\frac{5t^2}{\sqrt2}+\frac{t^4}{2}$$

$$t^4-\frac{5t^2}{\sqrt2}-4=0$$

I will leave the task of solving for quadratic equations as an exercise.

Related Question