A train at a speed od 90km/hr. Breaks are appliedso as to produce the uniform acceleration of -0.5m/sec2. Find the distance the train travelled befor it is brought back to rest.

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Solution

We know the equation
v^2=u^2-2as
v=final velocity
u=iniial velocity
a=accelaration
s= dusplacement
here v=0
u=25m/s 90km/h=90×5/8 m/s
a=-.5
0-225=2×-.5×s
s=225m

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A train at a speed od 90km/hr. Breaks are appliedso as to produce the uniform acceleration of -0.5m/sec2. Find the distance the train travelled befor it is brought back to rest.

We know the equation v^2=u^2-2as v=final velocity u=iniial velocity a=accelaration s= dusplacement here v=0 u=25m/s 90km/h=90×5/8 m/s a=-.5 0-225=2×-.5×s s=225m

We know the equation
v^2=u^2-2as
v=final velocity
u=iniial velocity
a=accelaration
s= dusplacement
here v=0
u=25m/s 90km/h=90×5/8 m/s
a=-.5
0-225=2×-.5×s
s=225m