Initial velocity of vehicle is #u# =# 90 kmh^-1 #
# color(red)(= 90 xx (1000m)/(3600s) = 25 ms^-1)#
After applying brakes, he can stop in Distance #s = 50 m = 50/1000 = 0.05 km#
#color(red)(=50 m )#
Final velocity #v = 0 kmh^-1# as the vehicle stops.
(a) To find the deceleration of the car:
Apply equation:
#s = (v^2 -u^2)/ (2a)#
#=> a = (v^2 -u^2)/ (2s)#
#a = (0^2 - 90^2)/ (2 xx 0.05)#
#color(red)( =(0-25^2)/ (2 xx 50))#
# a = -8100/ 0.1 = -81000 kmh^-2#
#color(red)(=-625/100 = -6.25ms^-2)#
#therefore #Deceleration = #81000 kmh^-2#
# OR color(red)(=>81000 xx (1000m)/ (3600s xx 3600s ) =6.25 ms^-2 = "deceleration")#
Note : We give negative sign if acceleration is to be written and positive sign if the word deceleration or retardation is to be written.
(2) To find the time elapsed before he stops:
Apply equation :
# v= u + at#
# => t = (v-u)/a # #(kmhr^-1)/ (kmhr^-2)#
# t= (0 -90)/ -81000#
#t = 0.00bar111 hr. = 0. 0bar6# minutes = # 4
#seconds#color(red)(= -25/-6.25 = 4s)#
#therefore # a)#81000 kmh^-2 = color(red)(6.25ms^-2)# is the deceleration #color(red)(OR)# #color(red)("accceleration" = - 6.25 ms^-2)#of the car
and
b) #4# seconds is the time elapsed before he stops.
