surely you mean 100 m not 100 km
and 40 m/s not km/s (40,000 meters per second is quite a train)
and what is the final speed ?
v = Vi + a t
d = Vi t + (1/2)a t^2
v = 40 + a t ...... whatever v is so a = (v-40)/t
( a is negative of course)
use that in
100 = 40 t + (1/2) a t^2
solve for t and go back and get a
then do for v = 0 to find stopping distance
The velocity of a train is reduced uniformly from 40km/sec while travelling a distance of 100km.
A. Calcukate the retardation.
B. How much farther will it trqvel before coming to rest, assuming the same retardation.
3 answers
First, do you mean 40km/sec? That's one fast train.
Maybe you meant 40 m/s. That's about 144 km/hr. More likely.
Or, even better, maybe 40 km/hr = 17.88 m/s
Now, the real problem is, you don't say how long it takes to cover that distance, nor what its velocity then was. So there's no way to find the acceleration. And no way to extrapolate to when v=0.
In any case, you just have to use
s(t) = vt + 1/2 at^2
to solve this problem.
Maybe you meant 40 m/s. That's about 144 km/hr. More likely.
Or, even better, maybe 40 km/hr = 17.88 m/s
Now, the real problem is, you don't say how long it takes to cover that distance, nor what its velocity then was. So there's no way to find the acceleration. And no way to extrapolate to when v=0.
In any case, you just have to use
s(t) = vt + 1/2 at^2
to solve this problem.
Ikr that's why I wasn't getting the answer. Thanks.
2 answers