# Homework Help: Calculating constant velocity, deceleration, and total displacement

1. Jun 24, 2012

### pamelajanas

1. The problem statement, all variables and given/known data

An object starts at rest and accelerates at 2.5 m/s^2 for 12 seconds. It then moves at a constant velocity for 5.0 seconds and then slows down uniformly at 1.5 m/s^2 until it stops. Calculate:
a) the constant velocity
b) the time taken for deceleration
c) the total displacement

2. Relevant equations

"possibly"
a=v/t or any of the five equations for uniform acceleration

3. The attempt at a solution
I know how to do part c) but i need the first two answers which i'm not sure how to use.
For a) i don't know which method to use: a=v/t or some kinematic equation, or which acceleration to use or which time (the 12 seconds while its accelerating, or the 5 seconds for constant velocity.
For b) i don't know how to do either because it i based on part a)

2. Jun 24, 2012

### sankalpmittal

Hi pamelajanas !!
Welcome to PF !!
:)

Hints :
Employ the first equation of 1-D (uniform acceleration) motion to solve for constant velocity, part (a). Also note that a = Δv/Δt and not v/t.
Consequently you will know which kinematic equation to apply to solve for (b) and (c).

Get on it !!

3. Jun 24, 2012

### pamelajanas

okay so using the first equation for part a) I would have to rearrange it in order to find v2?
meaning...
v1 = 0 m/s (which makes sense)
Δt= 12s or 5s (12s, which is the time for the acceleration or 5s, which is the time for the constant velocity)
Δd= .. i am not even sure

and i DID mean the a=Δv/Δt, which i do or do not have to use?

4. Jun 24, 2012

### pamelajanas

i'm still not sure how to do this.. :(
could you please look just above this, which are the reasons why i do not understand this

Last edited: Jun 24, 2012
5. Jun 25, 2012

### sankalpmittal

First equation of 1-D motion is :

v=u+at

You are to find v and you are given u and t. (The object starts from rest.)
To reach the velocity v via acceleration you're given time as 12 seconds.

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