The Student Room Group

A Level Mechanics.

In this question, the vectors i and j and unit vectors east and north respectively.
(a) A particle moves in two dimensions with constant acceleration. Initially it has
position vector 3i+2j m relative to a fixed origin and has initial velocity i+j ms-1.
After 4 seconds it has position vector 7i-4j m.
What is its velocity at that time? [4]
(b) The particle then continues at constant speed.
Calculate how long the particle has been travelling at constant speed when it is southeast of the origin. [4]

I did part a using s=1/2(u+v)t

For part b, I know that SE i and j components will be equal. How do I start this question?
Reply 1
Original post by Sasuto
In this question, the vectors i and j and unit vectors east and north respectively.
(a) A particle moves in two dimensions with constant acceleration. Initially it has
position vector 3i+2j m relative to a fixed origin and has initial velocity i+j ms-1.
After 4 seconds it has position vector 7i-4j m.
What is its velocity at that time? [4]
(b) The particle then continues at constant speed.
Calculate how long the particle has been travelling at constant speed when it is southeast of the origin. [4]

I did part a using s=1/2(u+v)t

For part b, I know that SE i and j components will be equal. How do I start this question?


SE, the position vector will be
xi - xj
so the j coefficient is negative and you want to find the time when this occurs. You have the position vector and (constant) velocity vector at 4s, so just use hte usual speed-distance-time formula to solve (simultaneously) for the time.
(edited 9 months ago)
Reply 2
Original post by mqb2766
SE, the position vector will be
xi - xj
so the j coefficient is negative and you want to find the time when this occurs. You have the position vector and (constant) velocity vector at 4s, so just use hte usual speed-distance-time formula to solve (simultaneously) for the time.

Thanks man!!
Reply 3
Original post by mqb2766
SE, the position vector will be
xi - xj
so the j coefficient is negative and you want to find the time when this occurs. You have the position vector and (constant) velocity vector at 4s, so just use hte usual speed-distance-time formula to solve (simultaneously) for the time.

How would you solve it simultaneously?
Reply 4
Original post by Jdhehebe
How would you solve it simultaneously?


Not sure what you think "it" is, it would be good to upload your thoughts. But you should set up a couple of equatoins to represent the motion in that phase and note that at a certain time, the position will be SE of the origin so the coordinate displacements are related. You could do a geometry or relative motion approach as well.
Reply 5
Original post by mqb2766
Not sure what you think "it" is, it would be good to upload your thoughts. But you should set up a couple of equatoins to represent the motion in that phase and note that at a certain time, the position will be SE of the origin so the coordinate displacements are related. You could do a geometry or relative motion approach as well.

I have the constant velocity and position vector points at 4 seconds. Velocity = i - 4j and position vector = 7i - 4j and when using the speed distance time equation, you get time= distance/speed. But how does that help as the position and velocity are in component form...?
Reply 6
The
distance = speed*time
applies to both the i and j directions seperately to give two simultaneous equations. Or it could be applied in the direction of motion. Its just a very simple suvat
s = ut + 1/2 at^2
where a=0 so its really not different.

Original post by Jdhehebe
I have the constant velocity and position vector points at 4 seconds. Velocity = i - 4j and position vector = 7i - 4j and when using the speed distance time equation, you get time= distance/speed. But how does that help as the position and velocity are in component form...?
Reply 7
Original post by mqb2766
The
distance = speed*time
applies to both the i and j directions seperately to give two simultaneous equations. Or it could be applied in the direction of motion. Its just a very simple suvat
s = ut + 1/2 at^2
where a=0 so its really not different.

Okay so using the Suvat method, I am confused with what values you would substitute in. I keep ending up with t=7 or t=1 when subbing in the values of position and velocity that I already have?
Reply 8
Original post by Jdhehebe
Okay so using the Suvat method, I am confused with what values you would substitute in. I keep ending up with t=7 or t=1 when subbing in the values of position and velocity that I already have?


Ive not worked it through so assuming your part a) is correct, then for this phase is
u = i - 4j
s0 = 7i - 4j
And you could/shouldsketch this so its currently at the point (7,-4) and the direction of motion (arrow) is (1,-4). It wants to intersect the line y=-x (SE). So as above you could do geometry or line intersection or using suvats you have in the x and y coordinates
x-7 = t
x-4 = 4t
so in each case distance from the initial point = speed*time, where t is the time from the initial point Also as above you could do it using relative motion using 1 speed/distance/time equation, but Id guess the question is leadinng you to the usuall x-y approach/
(edited 5 months ago)

Quick Reply

Latest