So basically I got this question and it's baffling. Over a distance of 5m the speed of a bicycle increase's from 2ms to 5ms what I's the magnitude of acceleration?

Answer 1

a = change in v/ change in time
= (5-2)/time
well how long did it take?
average speed = 7/2 =3.5
time = 5/3.5
so
a = 3 /(5/3.5)
= 2.1 m/s^2

Answer 2

You can use average speed with constant acceleration because
v = Vi + a t, straight line

Answer 3

Thanks. I understand it now. But we have like set formulas in our tables for incidence like v=u+at etc. I have learnt the changing velocity over time but in a test situation would I have to use the formulas given or the method appove?

Answer 4

distance from t1 to t2 =
integral v dt = Vi (t2-t1) + .5a (t2-t1)^2
distance/time
= average speed
= Vi + .5 a (t2-t1)
but speed at t2 is v2 = Vi + a (t2-t1)
so a(t2-t1) = v2-vi
so
average speed = Vi + .5(V2-Vi)
= .5 (Vi+V2)
if you really wanted to know :)

Answer 5

Relating to the first question. How long did it take to do this?

Answer 6

well to do the problem with your formulas
v = Vi + a t
5 = 2 + a t
so
3 = a t so t = 3/a

and
d = Vi t + .5 a t^2
5 = 2 t + .5 a t^2
5 = 2 (3/a) + .5 a (9/a^2)
5 = 6/a + 4.5/a
5 a = 10.5
a = 2.1 m/s^2 again

Answer 7

time needed is 3/a

Answer 8

Got the answer it's basically 5-2/2.1=1.43s