A car traveling at 25.0 m/s runs out of gas while traveling up a 19.0 ∘ slope.How far up the hill will it coast before starting to roll back down?

gas-while-traveling

#1

A car traveling at 25.0 m/s runs out of gas while traveling up a 19.0 ∘ slope.How far up the hill will it coast before starting to roll back down?

Answer:

U can use the conservation of energy

Kinetic Energy = 0.5mv^2 = 0.5*625m
now let it go up the height h , from conservation on energy , initial KE + initial PE = final KE + final PE
Thus inital PE and final KE = 0

0.5*625m=mgh
or 312.5=10h ( i am taking gravity as 10 m/s2)
h= 31.25 m

Now , let distance travelled by it = y

then ysin19=h
or y=h/sin19= 31.25/sin19 = 31.25 cosec19 = 208.504 meter

ask for any doubts :slight_smile:

on inclined surface force acting on it parellar to the plane of surface = -mgsin19
this acceleration = F/m = -gsin19
since v2=u2 + 2as
v= 0
u=25 m/s
thus we have , 0 = 625 - 20sin19s
or s= 625/20sin19 = 208.5 meters

ask for any furthur doubts