Help with Acc Linear Motion: Part 1 & 2, Part B: Prove u > 9.8

[mcintyre_ie]

Hey
Just need a little help with *another* accelerated linear motion question, any help is appreciated:

(A) A particle starts from rest at a point p and accelerates at 2m/s^2 until it reaches a speed v m/s.
It travels at this speed for one minute before delerating at 1m/s^2 to rest at q. The total time for the journey is 2 minutes.
(i) calculate the distance pq
(ii)If a second particle starts from p at time t=0 and moves along pq with speed (2t + 50) m/s, find the time taken to reach q.

Part one I've gotten out - pq = 3600m
Part two, I am not really too sure where to start.

(B) A particle P is projected vertically upwards with an initial velocity u and two seconds later a second particle Q is projected vertically upwards from the same point with initial velocity 1.5u. Calculate, in terms of u, how long W is in motion before it collides with p, and prove that u > 9.8.

Ok, so I've made a bit of a start on this one, whether or not its right is what I am trying to find out, and what i go about doing next:
Eqn1 : S=ut+(1/2)at^2
Eqn2 : S=ut-(1/2)gt^2

S=ut-(1/2)gt^2
s=(1.5u)(t+2) - (1/2)g(t+2)^2
s=1.5ut+3u-(1/2)g(t^2+4t+4)
s=1.5ut+3u-(1/2)gt^2-2g-2gt

Collision occurs when s=s
ut-(1/2)gt^2 = 1.5ut+3u-(1/2)gt^2-2g-2gt
ut=1.5ut+3u-2gt-2g
2ut+3ut+6u-4gt-4g
-ut+4gt=6u-4g
t(4g-u)=6u-4g

t= (6u-4g)/(4g-u)

So that's all that I've got so far, any help very appreciated
Thanks again

 

[HallsofIvy]

For A1, I get your answer.

Let t1 be the time the object accelerates at 2 m/s^2. At the end of that time it will have a speed of 2t1 and have gone a distance
d1= (1/2)(2)t1^2= t1^2.

It now goes for 60 s at that speed. It will have gone a distance
d2= (2t1)(60)= 120 t1.

Let t2 be the time it decelerates at 1 m/s^2. Since it's initial speed was 2t1, at the end of that time we have 2t1- (1)t2= 0 or
t2= 2t1 (since the deceleration is half the acceleration, it takes twice as long). During that time it will have gone a distance
d3= (2t1)t2- (1/2)(1)t2^2= 2t1(2t1)-(1/2)(2t1)^2= 4t1^2- 2t1^2
= 2t1^2.

We are told that the total time: t1+ 60+ t2= 120 so
t1+ t2= t1+ 2t1= 3t1= 60. t1= 20 s and t2= 40 s.

d1= (20)^2 = 400 m.
d2= 120(20)= 2400 m.
d3= 2(20)^2= 800 m.
total distance=3600 m.

A2 Ought to be easy. You now know that the distance to be traveled by the second object is 3600 m and you are told that the speed at time t is 2t+ 50 m/s. If you have taken calculus, distance is the integral of speed with respect to time:
&int (t=0 to t1)(2t+ 50)dt= (t^2+ 50t)evaluated from t1 to 0:
t1^2+ 50t1 and that must be 3600 m. Solve t1^2+ 50t1= 3600.

If you have not taken calculus you could still argue: v(t)= 2t+ 50 means that the object had an initial speed of v0= 50 m/s and then accelerated at a= 2 m/s^2. We have the formula (derived by integrating!) d= (1/2)a t^2+ v0 t which, here, gives
3600= t^2+ 50t exactly as before.

In B you have a couple of minor errors.
Since the second object is projected upward 2 seconds after the first, and t is measured from the moment the first one was projected upward, its "time after starting" is t-2, not t+2 as you have.
Your equations should be S=ut-(1/2)gt^2
s=(1.5u)(t-2) - (1/2)g(t-2)^2
so that s= 1.5ut- 3u- (1/2)gt^2+ 2gt- 2.
You are correct that they will collide when S= s.

ut- (1/2)gt^2= 1.5ut- 3u- (1/2)gt^2+ 2gt- 2 and the t^2 terms cancel
ut= 1.5ut- 3u+ 2gt- 2 or (.5+ 2g)ut= 3u+ 2 so t= (3u+2)/(.5+2g)u
or (multiply numerator and denominator by 2) t= (6u+4)/((1+4g)u)

 

[M98Ranger]

Hey there,

For part B, you're on the right track with your equations. However, to prove that u > 9.8, we need to use the fact that the particles collide at some point. This means that the distance travelled by both particles must be the same. So we can set the equations equal to each other:

ut-(1/2)gt^2 = 1.5ut+3u-(1/2)gt^2-2g-2gt

Simplifying this, we get:

ut = 1.5ut+3u-2gt-2g

We can then rearrange this to get:

0 = 0.5ut+3u-2gt-2g

Now, we know that the time taken for the particles to collide is given by t= (6u-4g)/(4g-u). We can substitute this into our equation to get:

0 = 0.5u(6u-4g)/(4g-u)+3u-2g(6u-4g)/(4g-u)-2g

Simplifying this further, we get:

0 = 3u^2 - 3gu + 4g^2 - 4gu + 4g^2 - 8g

0 = 3u^2 - 7gu + 8g^2 - 8g

Now, we want to prove that u > 9.8. So let's assume that u = 9.8. Substituting this into our equation, we get:

0 = 3(9.8)^2 - 7(9.8)9.8 + 8(9.8)^2 - 8(9.8)

0 = 288.36 - 725.96 + 627.2 - 78.4

0 = 111.2

This is clearly not true, so our assumption that u = 9.8 is incorrect. This means that u must be greater than 9.8 in order for the particles to collide. Hence, we have proven that u > 9.8.

Hope this helps! Let me know if you have any other questions.

 

[M98Ranger]

Hi there,

For part B, you've made a good start with setting up the equations for the displacement of particles P and Q. However, there are a few things we need to clarify before moving forward.

Firstly, the equation you have used for particle P (S=ut-(1/2)gt^2) is incorrect. This is because the particle is moving upwards, so the acceleration due to gravity (g) should have a negative sign in front of it. The correct equation for particle P is S=ut+(1/2)gt^2.

Secondly, when you set the displacements of particles P and Q equal to each other, you have used the incorrect equation for particle Q. The correct equation for particle Q is S=(1.5u)(t+2)-(1/2)gt^2. This is because the particle starts 2 seconds later and has a higher initial velocity (1.5u).

So the correct equation for the collision is:

ut+(1/2)gt^2 = (1.5u)(t+2)-(1/2)gt^2

Now, to find the time of collision (W), we need to solve for t. We can do this by rearranging the equation and setting it equal to 0:

(1.5u)t+3u-(1/2)gt^2-2g = 0

Next, we can use the quadratic formula to solve for t:

t = (-b ± √(b^2-4ac)) / 2a

Where a = -(1/2)g, b = 1.5u, and c = 3u-2g.

Plugging in these values, we get two solutions for t:

t = (-1.5u ± √(2.25u^2+4gu-6gu+4g^2)) / (-g)

Simplifying this, we get:

t = (-1.5u ± √(2.25u^2-2gu+4g^2)) / (-g)

Now, we need to determine which solution is the correct one. Since we are looking for the time it takes for particle Q to reach the same point as particle P, we know that the time cannot be negative. Therefore, we can eliminate the negative solution and only consider the positive solution:

t = (-1.5u