VCE Specialist 3/4 Question Thread!

[Shiney94]

Could anyone please help me with this question:
A car is accelerating at a constant rate from rest, it then reaches a speed of 60km/h after travelling 50 m.
At what speed will the car be travelling at the end of the 11th second?

[Phy124]

Could anyone please help me with this question:
A car is accelerating at a constant rate from rest, it then reaches a speed of 60km/h after travelling 50 m.
At what speed will the car be travelling at the end of the 11th second?

[Shiney94]

Thanks for that. Just another question,
A car travels at a constant velocity of 75km/h, passing a stationary police car which immediately sets off in pursuit at a constant rate of acceleration. It takes the police car 80 seconds to catch the car, find:
a) The acceleration of the police car
b) The speed of the police car when the other car has been reached.

[Jenny_2108]

Thanks for that. Just another question,
A car travels at a constant velocity of 75km/h, passing a stationary police car which immediately sets off in pursuit at a constant rate of acceleration. It takes the police car 80 seconds to catch the car, find:
a) The acceleration of the police car
b) The speed of the police car when the other car has been reached.

75km/h= 125/6 m/s
When the police car catches the car, the distances are both the same
At t=80s, distance of the car= 80. 125/6=5000/3 (m)
Let v= velocity of the police car when catching the car
5000/3=1/2.80.v => v=125/3 (m/s)
a. Acceleration of the police car = 125/3:80=25/48 (m/s2)
b. Speed of the police car=125/3 (m/s)

Correct me if I'm wrong

[soccerboi]

I need help with this question:
A particle travels in a straight line with a constant velocity of 4m/s for 12 seconds.It is then subjected to a constant acceleration in the opposite direction for 20 seconds which returns the particle to its original position.Find:
a)the acceleration of the particle

[Phy124]

I need help with this question:
A particle travels in a straight line with a constant velocity of 4m/s for 12 seconds.It is then subjected to a constant acceleration in the opposite direction for 20 seconds which returns the particle to its original position.Find:
a)the acceleration of the particle

Original displacement;



To return its original position;







i.e. 0.64 m/s² in the opposite direction to the original motion, given the original motion is defined as positive (which I did)

edit: Though I'd just mention you can also define certain variables as positive or negative, when told which direction they are in and it will just give you the magnitude.

[soccerboi]

^ The answer is somehow -0.64m/s²

Thanks!

[soccerboi]

I need help with this question:
A particle travels in a straight line with a constant velocity of 4m/s for 12 seconds.It is then subjected to a constant acceleration in the opposite direction for 20 seconds which returns the particle to its original position.Find:
a)the acceleration of the particle

How would i find the time the particle is traveling back towards its initial position? I thought it was just 12+20=32 s, but the answer is 55/4 s

[Bhootnike]

I need help with this question:
A particle travels in a straight line with a constant velocity of 4m/s for 12 seconds.It is then subjected to a constant acceleration in the opposite direction for 20 seconds which returns the particle to its original position.Find:
a)the acceleration of the particle

How would i find the time the particle is traveling back towards its initial position? I thought it was just 12+20=32 s, but the answer is 55/4 s

can you please rephrase that -  doesnt make any sense to me haha, or give the question ?
from what you say, it sounds like the question was the time the particle takes to get back to its initial position, which would just be 20 seconds cause thats given..
but i dont think thats the question!

[soccerboi]

I need help with this question:
A particle travels in a straight line with a constant velocity of 4m/s for 12 seconds.It is then subjected to a constant acceleration in the opposite direction for 20 seconds which returns the particle to its original position.Find:
a)the acceleration of the particle

How would i find the time the particle is traveling back towards its initial position? I thought it was just 12+20=32 s, but the answer is 55/4 s

can you please rephrase that -  doesnt make any sense to me haha, or give the question ?
from what you say, it sounds like the question was the time the particle takes to get back to its initial position, which would just be 20 seconds cause thats given..
but i dont think thats the question!

Part b) Find the time the particle is travelling back towards its original position.

[b^3]

Firstly you need to find when it changes direction, i.e. when it comes to rest.
So taking when the acceleration is first applied as







So the total time it was decelerating was 20 seconds, and we know that after 6.25 seconds it changed direction, so the time left is the time that it was travelling back to its starting position.
So that is

[paulsterio]

Hmm, are you sure b^3, I personally think the question is asking when it changes direction?

i.e. 25/4 s

[b^3]

Hmm, are you sure b^3, I personally think the question is asking when it changes direction?

i.e. 25/4 s

Thats the way I originally read it, but the OP did say the answer was 55/4 seconds.

I need help with this question:
A particle travels in a straight line with a constant velocity of 4m/s for 12 seconds.It is then subjected to a constant acceleration in the opposite direction for 20 seconds which returns the particle to its original position.Find:
a)the acceleration of the particle

How would i find the time the particle is traveling back towards its initial position? I thought it was just 12+20=32 s, but the answer is 55/4 s

Its not a well worded question.

[paulsterio]

I think it should be worded: At what time does the particle REACH its initial position?

But OK

[b^3]

I think it should be worded: At what time does the particle REACH its initial position?

But OK

For that the answer would be 12+20=32 seconds wouldn't it?, as you need to include the first leg before the acceleration as well, and then still get back to the starting position, not where it turns around. It should be something along the lines of "After coming to rest, how long does it take for the particle to return to its original position?" As that specifies that only the time taken for the return journey is needed.

Probably looking into this too much but anyway...