A car starts from rest and accelerates uniform upto speed 90 km/hrs in 5 sec. Total distance covered by car in this time will be

• v=u+at which give a=5
  s=ut+1/2at^2 which give s=62.5m ans
• 10 years agoHelpfull: Yes(9) No(4)
• speed=90km/hr=90*5/18=25m/sec
  time =5 sec
  distance= speed*time
  =25*5=125m
• 9 years agoHelpfull: Yes(8) No(2)
• d=((v0+v)/2)*t
  d=((0+90)/2)*5)*5/18=62.5m ans
  see acceleration formula in wikipedia
• 9 years agoHelpfull: Yes(1) No(0)
• Speed of car at last instant of given interval
  = 90*5/ 18 m/s = 25 m/s
  so average speed = 0+25 / 2
  Therefore distance = 25/2 * 5 = 125/2 = 62.5 m
• 9 years agoHelpfull: Yes(1) No(0)
• Ans>77.7meters
  90/5=18
  18km/hr for 1st sec and 36,54,72 and 90km/hr at 5th sec
  average speed = 58km/hri.e 15.5m/sec
  so for 5 sec 77.7meters
• 9 years agoHelpfull: Yes(0) No(2)
• d= s *t
  90*(5/60) = 2.5 km
• 9 years agoHelpfull: Yes(0) No(1)
• speed is 90-0=90 kmph
  distance=speed* time=90*(5/18)*5=125 metre.
• 7 years agoHelpfull: Yes(0) No(0)