Question #20313

Train breaks are applied it has covered covered 800 m distance reducing speed from 96km/h to 48km/h before going going to rest how much distance it covered.

Expert's answer

From the statement we can get acceleration in time of reducing speed:

v0-at2=v1

v1=at1

at1=48

at2=48

acceleration is constant, therefore: t1=t2

So,

S1=v0t1-at1^2/2

S2=v1t1-at1^2/2

S2=S1-(v0-v1)t1

Now, we have to find t1

t1=48/a

S=v048/a-a48^2/a^2/2

S/48^2=1.5/a

So, 800/1.5*48^2=a

a=4.32

t1=48/4.32

S2=800-48*48/4.32=266.(6) m

v0-at2=v1

v1=at1

at1=48

at2=48

acceleration is constant, therefore: t1=t2

So,

S1=v0t1-at1^2/2

S2=v1t1-at1^2/2

S2=S1-(v0-v1)t1

Now, we have to find t1

t1=48/a

S=v048/a-a48^2/a^2/2

S/48^2=1.5/a

So, 800/1.5*48^2=a

a=4.32

t1=48/4.32

S2=800-48*48/4.32=266.(6) m

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