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

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

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.