# I dont understand what uniform acceleration is?

1. Sep 8, 2004

### faisal

hello fellow Physicians, My names Faisal and i'm a a physics student, i'm really stuck on a question & i could really use u'r help
Ok
An object with an initial speed of 3m/s begins to accelerate uniformly at 2m/s2. what will b its speed 5s later?

i dont understand what uniform acceleration is?
i really need to know the basics of physics before i can start anything advanced can u physicans recommend a good website?

Last edited by a moderator: Sep 15, 2015
2. Sep 8, 2004

### Alkatran

First of all, write down what you knwo and what you don't:
d (distance) = ?
vi (starting speed) = 3 m/s
vf (final speed) = ?
a (acceleration) = 2 m/s^2
t (time) = 5 s

Now, there are about 5 formulas that relate those 5 values. I can't name them off, but look at it logically: the acceleration is change of speed per second. The change of speed PER SECOND IS TWO, and it lasts FIVE SECONDS. So if you are going 2 faster every second, for 5 seconds, and you start at 3... what do you do?

Still need more help?

3. Sep 8, 2004

### poolwin2001

Velocity/speed is the rate of change of position.
Acceleration measures the rate of change of velocity.
Uniform acceleration ==>acceleration remains constant with time

For your Problem:
Remember these fudamental eqns:
$$(i)v=u + at$$

$$(ii)v^2=u^2 + 2as$$

$$(iii)s=ut+\frac{1}{2}at^2$$
Try to figure these out yourself
EDIT:where v is final velocity
u is initial velocity
a is acceleration
t is time taken
s is the displacement

Last edited: Sep 8, 2004
4. Sep 8, 2004

### marlon

Poolwin2001,

Do please give constructive answers. By just reciting formula's without defining what is what, you only make it more difficult.

regards
marlon

5. Sep 8, 2004

### humanino

Welcome Faisal ! You should use the specific help forum next time.
Enjoy PF !

6. Sep 8, 2004

### faisal

''An object with an initial speed of 3m/s begins to accelerate uniformly at 2m/s2. wh

i still dont understand the only equation which relates to this question
is accelertaion, this is what i originally worked out 2-3/5=0.68 the answer was obviously wrong.

7. Sep 8, 2004

### marlon

Use v = v_0 + at

v : speed
v_0 initial speed
a : acceleration

so you have v = 3 + 2*5 = 13 m/s

8. Sep 8, 2004

### Nenad

you have to understand what acceleration is first if you want to understand the problem properly. Dont worrry if you dont get it at first, a LOT of people do not get it untill after they work with the concept for about a month. Uniform Acceleration is the rate of change in speed, this means that every time one second passes by, the speed changes uniformly by a margin, it either decreases by that speed or increases. In your case, every second, the speed of the object increases by $$2m/s^2$$. You begin at $$3m/s^2$$ and you travel for 5 seconds. If we break the calculation down we would see the separate speeds at each second:
$$V_i = 3m/s$$ (no time has passed)
$$V_1 = 3m/s + 2m/s$$ (1 second has passed)
$$V_2 = 3m/s + 2m/s + 2m/s$$ (2 seconds have passed)
$$V_3 = 3m/s + 2m/s + 2m/s + 2m/s$$ (3 seconds have passed)
$$V_4 = 3m/s + 2m/s + 2m/s + 2m/s + 2m/s$$ (4 seconds have passed)
$$V_5 = 3m/s + 2m/s + 2m/s + 2m/s + 2m/s + 2m/s$$ (5 seconds have passed)

as you can see, the general pattern for the speed after a time $$t$$ and at constant acceleration $$a$$ and an initial speed $$V_o$$ is:
$$V = V_o + at$$

if you still dont understand, try looking up some acceleration problems and try doing them, doing practice probalems is the best way to learn.

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