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secondary 4 | A Maths
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LockB
LockB

secondary 4 chevron_right A Maths chevron_right Singapore

need help with part b, the answer is supposed to be 15.4 but i got the above answer

Date Posted: 2 years ago
Views: 245
J
J
2 years ago
Later I'll explain
J
J
2 years ago
v = 10 - 20/(3t + 1)²

What you are asked to find is distance travelled.

But integrating v with respect to t gives you displacement (which is the shortest distance between the starting point and the location of the particle at time t)

A negative displacement means you have
travelled some distance backwards.


What you realise is that the velocity is negative at first

(when t = 0, v = 10 - 20/(3(0)+ 1)²
= 10 - 20
= -10

This means the particle was travelling backwards at first.
LockB
LockB
2 years ago
sorry! i still dont understand how to find distance travelled... dont really understand how the whole thing works too
J
J
2 years ago
So what you should do is to find the point where v = 0 first.

10 - 20/(3t + 1)² = 0

10 = 20/(3t + 1)²

10(3t + 1)² = 20

(3t + 1)² = 2

3t + 1 = ± √2

3t = √2 - 1
or 3t = -√2 - 1

t = ⅓(√2 - 1)


or t = ⅓(-√2 - 1)

(rejected as this is negative, and time cannot be negative)


Now since this is the distance travelled backwards,


You got to do ∫ v dt from t = 0 to t = ⅓(√2 - 1) and put the modulus sign on the result

(as the displacement will come out as negative but you know that distance is a positive value)

You will get ≈ 0.57191

Then, do ∫ v dt from t = ⅓(√2 - 1) to t = 2

You will get ≈ 14.8576


Add the two and you'll get your ≈ 15.4
J
J
2 years ago
Basically from t = 0 to t = ⅓(√2 - 1), your particle is travelling backwards.

The velocity (remember that it is a vector magnitude, with direction) is negative.

Eg. When I say I'm travelling at -10km/hr, my velocity is -10km/hr but my speed is 10km/hr, just that I'm travelling in the backward direction

Speed has no direction.
For velocity, direction matters.



So here,

Your displacement is negative (-0.5719) , which means that the particle is now 0.5719 m behind its starting point.

But you know distance is a positive value so distance travelled here ≈ 0.5719
(you do a modulus/take its absolute value)

When it is t = ⅓(√2 - 1), the particle has stopped
moving backwards and has come to a stop (v = 0)

From this timing , it now starts to move forward.

So from this time point to t = 2, displacement is positive and there is no need to modulus to get the distance travelled.
J
J
2 years ago
No.

t = 0 is when the particle starts moving from point O.

It moves backwards.

t = ⅓(√2 - 1) is when it stops moving backwards and has come to a halt. Here, the velocity is zero (v = 0). The particle is at rest.


After t = ⅓(√2 - 1), it starts moving forwards
J
J
2 years ago
The particle is not like a car. It doesn't have a front or back end. It doesn't need to turn around.

It simply just stops and moves in the opposite direction.

Think of it like an MRT train (even though it has a back and front end, it doesn't do any uturns)


You're not familiar with velocity? Do you take any physics?
LockB
LockB
2 years ago
nope im taking pure biology so this chapter is kind of confusing to me
J
J
2 years ago
I see. Maybe you'll want to read up online on what velocity, displacement and acceleration are.
LockB
LockB
2 years ago
for this kind of question, the first step to do is to check if the velocity is positive or negative when t=0? if it is positive means it is travelling forward and the question can be solved by finding the displacement

if it is negative,it is travelling backwards first and we must find the time where the particle halts (v=0) to know when it stops moving backwards and start to move forward. then continue solving by finding the displacement

is this correct
J
J
2 years ago
Largely correct.

However, sometimes it is the opposite.

The particle may be travelling forward at first.

Then, it may slow down, stop, and then start to travel backwards.

Aways check the value of t when v = 0.

And check if v is positive or negative on either side of that value of t.
J
J
2 years ago
Eg.

Let's say from t = 0 to t = 2, it is travelling forwards.

From t = 2 to t = 3, it is travelling backwards.

If you do ∫ v dt from t = 0 to t = 3 straight away, you will get the wrong answer for distance travelled.

You'll need to do integrate from t = 0 to t = 2 (it will be positive)

Then from t = 2 to t = 3, the value will be negative so if you're finding distance travelled, you'll need to modulus it / take the absolute value.

Then add the two up.


Sometimes , if the question is difficult or tedious, you might have more than 1 time interval where the particle is travelling backwards (where v is negative)

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LockB
LockB
2 years ago
thx :) i am taking GCE O Level