A car from rest accelerates uniformly & passes two points 120m apart in 6.0s. If the car passes second point with velocity 25.0m/s, what is velocity when it passes the first point? what is the distance of the second point from the starting point?

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Answer 1 · 2018-06-20T16:00:33

The velocity is #=15ms^-1#. The distance is #=187.5m#

Apply the equations of motion

#v=u+at#

and

#s=ut+1/2at^2#

#(i)#

The distance is #s=120m#

The time is #t=6s#

The velocity is #v=25ms^-1#

The velocity as it passes the first point is #u_1#

Therefore,

#25=u_1+a6#

#120=u_1*6+1/2a*6^2#

Solving the #2# equations

#{(25=u_1+6a),(120=6u_1+18a):}#

#<=>#, #{(u_1+6a=25),(u_1+3a=20):}#

#<=>#, #{(a=5/3ms^-2),(u_1=15ms^-1):}#

#(ii)#

Applying the #2# equations

#15=0+5/3*t_1#

#=>#, #t_1=9s#

The distance is

#s_1=0*9+1/2*5/3*9^2=67.5m#

The distance from the starting point is

#d=s+s_1=120+67.5=187.5m#