Question with acceleration vector

[goonking]

Homework Statement

http://imgur.com/LravIr3

Homework Equations

The Attempt at a Solution

We know A is wrong, because a object going in a circle has acceleration (i'm not sure why that is, maybe someone can explain)

B is wrong because if a car in reverse is slowing down, it technically has positive acceleration, right?

D is wrong because refer to my explanation for B.

C and E, I have no idea what that formula is but V x T is distance but I have no idea what the formulas are implying. Anyone can shed some light on what C and E means?

 

[Suraj M]

The formula $v(T + Δ T)$ refers to the velocity of the object at time$(T +Δ T)$.
Now can you decide if C and E are right or wrong?

 

[goonking]

The formula $v(T + Δ T)$ refers to the velocity of the object at time$(T +Δ T)$.
Now can you decide if C and E are right or wrong?

how does velocity multiplying the time give you velocity again?

 

[Suraj M]

Its not multiplication, It's a way of representing the velocity at a particular time, as the velocity varies with time!

 

[goonking]

Its not multiplication, It's a way of representing the velocity at a particular time, as the velocity varies with time!

can you make up a word problem that uses v(T+ΔT)?

 

[Suraj M]

Okay instead of a word problem with $V(T+ΔT)$ try this.
Let $~~ V(t) = ƒ(t)$ and $ƒ(t) = u(0) + at$
here i used $u(0)$ itts actually $u(t=0)$ we often miss out the $t=$
so in your question it should have been- $v(t_2 = T+ΔT)$ and $v(t_1 = T)$
we often omit the t= to make things easier,
Its represented like this because velocity is not a constant and is a function of time, skips the steps you'd have to involve to define $v_1 ~ and~ v_2$
its just like writing $V(initial) and V(final)$
see this

 

[NascentOxygen]

answer (D) seems true
EDIT: (D) is not always true

 

[NascentOxygen]

can you make up a word problem that uses v(T+ΔT)?

Where the acceleration of a body is known to be constant, we can determine that acceleration by making two measurements of its velocity some time apart, then using the formula;

a = Δv / Δt

= ( v(T+ΔT) - v(T) ) / ΔT

 

[goonking]

answer (D) seems true

if I'm in a car, and I'm in reverse while increasing my speed. I'm accelerating while going backwards so my acceleration is negative. I then slow down a bit while reversing but I'm still moving backwards, my acceleration is now positive even though I'm still moving backwards. It is positive because I slowed down in reverse. How can D still be true?

 

[NascentOxygen]

if I'm in a car, and I'm in reverse while increasing my speed. I'm accelerating while going backwards so my acceleration is negative. I then slow down a bit while reversing but I'm still moving backwards, my acceleration is now positive even though I'm still moving backwards. It is positive because I slowed down in reverse. How can D still be true?

You are right. I was not correct because in that case (D) is not true. I'll amend my earlier post. Thanks for the correction.