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###### A Level Mechanics.

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5 months ago

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?

(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?

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?

(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 5 months ago)

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.

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!!

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.

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?

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.

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...?

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.

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...?

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.

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?

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 1 month ago)

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