+11912 A train is travelling at a speed of 90km h . Brakes are applied so as to produce a uniform acceleration of= 0.5m/s^2 . Find how far the train will go before it is brought To rest.
(Explain in detail using the most simplest English you know with a perfect grammar)( or else .....just give the answer ....don't even bother about the English)
This equation maybe able to help you!.
V²=U²+2AS,
WHERE V=FINAL VELOCITY
U=INITIAL VELOCITY
A=ACCELERATION
S=DISTANCE OF ACCELERATION
+129852 I'm not great at Physics, Rosala, but I'll try....if the brakes are applied, the acceleration must be negative....since the train is slowing down!!!....so we have
(Vf)^2 = ( Vi )^2 + 2AS where Vf = the final velocity [0], Vi is the initial velocity[90km/h] , A is the acceleration = [-0.5m/s^2] and S is the displacement [stopping distance] ....so we have
0^2 = (90km/hr)^2 + 2(-0.5m/s^2)*S
But.....so that the units are the same, we must convert the 90km/hr to m/s......since there are 3600 seconds in one hr and 1000m in 1 Km....we have
90km/hr x 1 hr / 3600 sec x 1000m/ 1 km = 25m/s
And (25m/s)^2 = 625m^2/s^2
Now we have ..... [0^2 = 0]
0 = 625m^2/s^2 + 2(-0.5m/s^2)*S subtract the 625m^2/s^2 from both sides ................. [ and 2 * -0.5 = -1]
-625m^2/s^2 = -1m/s^2 * S multiply through by s^2
-625m^2 = -1m *S divide through by -1m
[-625m^2] / [ -1m] = S = 625m to stop the train
[ I think this is correct..... I hope !!! ]
