MedVision ad

Differentiation Question Help (1 Viewer)

202025

New Member
Joined
May 12, 2018
Messages
5
Gender
Male
HSC
2020
A body starts from rest and moves in a straight line so that its velocity v ms^-1 after t seconds is given by v = 2t + 6t^2. Calculate:

(a) its acceleration at the end of the first second

(b) its displacement after 5 seconds, given that the body is initially at zero displacement.

I have solved (a), however I am struggling with (b)


note: This question is from the differentiation chapter in my textbook therefore I believe that a method other than integration should be used.
 

HoldingOn

Active Member
Joined
Dec 18, 2016
Messages
318
Location
The Cosmos
Gender
Male
HSC
2018
I think they want you to use integration hence they have specified an initial condition for the body's displacement- indicating that the constant is zero. Also note that the acceleration is not uniform as the velocity function forms a parabola- so we can't use shape formulas like area of a triangle.
 
Last edited:

HeroWise

Active Member
Joined
Dec 8, 2017
Messages
353
Gender
Male
HSC
2020
Sketch velocity and find area? it woul;d be an approximate though.


isnt primitive functions in differentiation chap?
 

202025

New Member
Joined
May 12, 2018
Messages
5
Gender
Male
HSC
2020
This question is from Chapter 7 - Calculus - Introduction to Differentiation; from the textbook New Senior Mathematics Advanced Year 11&12, Third Edition. Primitive functions are not introduced until Chapter 16 - The anti-derivative.
 

HoldingOn

Active Member
Joined
Dec 18, 2016
Messages
318
Location
The Cosmos
Gender
Male
HSC
2018
This question is from Chapter 7 - Calculus - Introduction to Differentiation; from the textbook New Senior Mathematics Advanced Year 11&12, Third Edition. Primitive functions are not introduced until Chapter 16 - The anti-derivative.
The only way I can see how to do it without integration is taking the average velocity and then multiplying by the time. But this would only give you an approximation.
 
Last edited:

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top