# Two dimensional motion with variable acceleration

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.

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

$$a(t)=2F(t)$$ 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.

• Thanks. I've managed to get the answer. However I'm still confused about the objects moving in the same direction. Are the velocities of the two objects always equal as you have said? Can't they still be travelling in the same direction with different speeds that may not always be equal. For example I could walk next to someone and walk at a different speed but still be travelling in the same direction. Also, I'm not sure how the technique of dividing the components by each other works. Is that the only way to solve for t as I've tried other things and didn't get very far. – Ben Harris Apr 6 at 21:00
• I did indeed skip some steps. Suppose the direction is $v_xi+v_yj$ and $w_xi+w_yj$. Suppose they are in the same direction, then $v_xi+v_yj=k(w_xi+w_yj)$ for positive $k$. that is we have $v_x=kw_x$ and $v_y=kw_y$. For this question, it is clear that $v_x$ and $w_x$ are both positive (they should be in the same sign), hence we have $\frac{v_y}{v_x}=\frac{kw_y}{kw_x}=\frac{w_y}{w_x}$, this gives us an equation in $t$. – Siong Thye Goh Apr 7 at 2:18
• Thanks! ${}{}{}{}{}$ – Ben Harris Apr 7 at 14:34