acceleration?

Question

the speed of a train is reduced uniformly from 10 m/s to 5 m/s while covering a distance of 150 m. what is its acceleration?

i need an explanation and a formula..
thanks!!

Answer 1

you are probably familiar with the two more common formulae liking distance (S), intial velocity (u), final velocity (v), time (t) and acc (a), these being:

1. s = ut + 1/2 a t^2
2. v = u + at

in this problem we dont have t, but we have s, u, and v and need to find a.

so, we need to eliminate "t" from formulae 1 and 2
from 2, we get t = (v-u)/a

substitute in formula 1:
S= u(v-u)/a + 1/2 a (v-u)^2 / a^2

multiply everything by 2a

2as = 2u(v-u) + (v-u)^2
2as = 2uv - 2u^2 + v^2 - 2uv +u^2

cancelling, you get 2as = v^2 - u^2

Answer 2

These are all the kinematic equations you will ever need.

X = T *(Vo + V)/2
X = Vo * T + 1/2 * A * T^2
V = Vo + A * T
V^2 = Vo^2 + 2 * A * X

You can just use the last one to solve this. 5*5 = 10*10 +2*A*150 => 25 = 100 + 300*A => A = -75/300 = -0.25 m/s/s

Answer 3

According to newton's third equation



v=5m/s
u=10m/s
s=150m

v^2=u^2+2as
5^2=10^2+2*a*150
25=100+2*a*150
a=(25-100)/300
= -0.25
acceleration is always positive
hence,acceleration is 0.25m/s^2

Answer 4

So s= v0*t +0.5a*t^2, v1=v0 +a*t, being v0=10, v1=5, s=150, t is time to cover s=150m, ‘a’ to be found; hence t=(v1-v0)/a, s= v0(v1-v0)/a +0.5a(v1-v0)^2/a^2;
Now as = (v1-v0){v0 +0.5v1 –0.5v0} =
= 0.5(v1-v0)(v1+v0) = 0.5(v1^2 -v0^2);
And a = 0.5(v1^2 -v0^2)/s = -0.25m/s^2

Answer 5

Vf²=2ad+Vi²

(5m/s)² = 2a(150m) + (10m/s)²

25m²/s²= 300m(a) + 100m²/s²

-75m²/s²= 300m(a)

a= -.25m/s²

Answer 6

v^2=u^2-2as