a car is moving at a rate of 90 km per hour and applies brakes which provide a retardation of 6 m per second how much time does the car takes to stop how much distance does the car cover before coming to rest what would be the stopping distance needed if speed of the car is doubled
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To answer this question use the formula: v=u+ats
={(u+v)/2}tv2
=u2+2as s
=ut+1/2at2 where s-displacement, v-final velocity, u-initial velocity, a-acceleration/retardation, t-timei)
How much time does the car take to stop?
in this case V=90 km/hr (convert to m/s=25m/s), u=0m/s, a=m/s2
since v=u+at...at=v-u....t=(v-u)/a
therefore t=(25-0)/6.....=25/6
=4.16s
therefore it takes 4.16 seconds to stop
ii)distance covered before stopping: displacement (s)
s= [(u+v)/2]t....
=0+25/2 x 4.16....
= 12.5 x 4.16= 52m
shence distance covered is 52 meters
iii)what is stopping distance if speed is doubled?
that means v= 52m/shence displacement s,s= [(u+v)/2]t....= 52/2 x 4.16 displacement therefore at double speed = 108.16 meters
={(u+v)/2}tv2
=u2+2as s
=ut+1/2at2 where s-displacement, v-final velocity, u-initial velocity, a-acceleration/retardation, t-timei)
How much time does the car take to stop?
in this case V=90 km/hr (convert to m/s=25m/s), u=0m/s, a=m/s2
since v=u+at...at=v-u....t=(v-u)/a
therefore t=(25-0)/6.....=25/6
=4.16s
therefore it takes 4.16 seconds to stop
ii)distance covered before stopping: displacement (s)
s= [(u+v)/2]t....
=0+25/2 x 4.16....
= 12.5 x 4.16= 52m
shence distance covered is 52 meters
iii)what is stopping distance if speed is doubled?
that means v= 52m/shence displacement s,s= [(u+v)/2]t....= 52/2 x 4.16 displacement therefore at double speed = 108.16 meters
abhaypawar:
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