A train is moving with a velocity of 90 km/h. It is brought to stop by applying the brakes wch produce a retardation of 0.5m/s square find the velocity after ten second
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First, there is a trouble with the units. kilometers and hours need to be translated into meters and seconds.
1 km/hr = 1000 meters / 3600 s = 0.2777777… m/s
So 90 km/hr = 90 x 0.277777777.. m/s = 25 m/s
Okay, now the mathematics should be easy. Just use the formula for velocity v and acceleration a over time t:
v = a t
If I am going 25 m/s, and slow down 0.5 m/s² for 10 s, that’s a total of 0.5 x 10 m/s of velocity lost. 5 m/s, to put it simply.
So, if I was going 25 m/s and slowed down by 5 m/s, I’d still be going 20 m/s.
Piece of cake, but let’s convert that back to km/hr just to make the answer fit more neatly with the problem:
20 m/s / 0.2777777777777… m/s/km/hr = 72 km/hr
To come to rest, the train must decelerate to 0 m/s. So, we will just solve our equation for t, instead of solving for v. Rearrange the equation by dividing both sides by a.
t = v / a
25 m/s divided by 0.5 m/s² should tell us the total time it takes to come to rest, then.
t = 25 m/s / 0.5 m/s² = 50 s.
So, it will take 50 seconds for the train, initially moving at a velocity of 90 km/hr, to come to rest.
1 km/hr = 1000 meters / 3600 s = 0.2777777… m/s
So 90 km/hr = 90 x 0.277777777.. m/s = 25 m/s
Okay, now the mathematics should be easy. Just use the formula for velocity v and acceleration a over time t:
v = a t
If I am going 25 m/s, and slow down 0.5 m/s² for 10 s, that’s a total of 0.5 x 10 m/s of velocity lost. 5 m/s, to put it simply.
So, if I was going 25 m/s and slowed down by 5 m/s, I’d still be going 20 m/s.
Piece of cake, but let’s convert that back to km/hr just to make the answer fit more neatly with the problem:
20 m/s / 0.2777777777777… m/s/km/hr = 72 km/hr
To come to rest, the train must decelerate to 0 m/s. So, we will just solve our equation for t, instead of solving for v. Rearrange the equation by dividing both sides by a.
t = v / a
25 m/s divided by 0.5 m/s² should tell us the total time it takes to come to rest, then.
t = 25 m/s / 0.5 m/s² = 50 s.
So, it will take 50 seconds for the train, initially moving at a velocity of 90 km/hr, to come to rest.
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U=90km/h = 90×5/18 = 25m/sec
v= 0m/s
a = -0.5m/s²
then using first equation of motion ,
v=u+at
0 = 25+(-0.5)t
0=25-0.5t
0.5t=25
t= 25/0.5 = 50 seconds
hope this helps
v= 0m/s
a = -0.5m/s²
then using first equation of motion ,
v=u+at
0 = 25+(-0.5)t
0=25-0.5t
0.5t=25
t= 25/0.5 = 50 seconds
hope this helps
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