A train accelerates at #-1.5# #m##/##s^2# for #10# seconds. If the train had an initial speed of #32# #m##/##s#, what is its new speed?

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Answer 1 · 2017-10-29T06:24:10

The new sped is #=17ms^-1#

Apply the equation of motion

#v=u+at#

The initial speed is #u=32ms^-1#

The acceleration is #a=-1.5ms^-2#

The time is #t=10s#

The final speed is

#v=u+at=32-1.5*10=32-15=17ms^-1#

Answer 2 · 2017-10-29T13:28:41

#17m/s#

We can solve this by using the kinematic equation

#color(blue)(v=u+at#

Where,

#color(orange)(rarrv="final speed"#

#color(orange)(rarru="initial speed"#

#color(orange)(rarra="acceleration"#

#color(orange)(rarrt="time taken"#

As know #3# of the values, we can apply them

#rarrv=32+(-1.5)(10)#

#rarrv=32+(-15)#

#rarrv=32-15#

#color(green)(rArrv=17 m/s#

So, the train attained a speed of #17m/s#

Hope this helps....God bless ☻