Question

Deceleration=4.5 m/sec² stopping distance=50metres velocity of car=......km/hr

Answer 1

heya friend ,


Applying $3^{rd}$  equation of motion

$v^{2}=u^{2}+2as$

[tex] v^{2} = u^{2}+2as [/tex]

$0= u^{2} - 2*4.5*50$

[tex]u^{2}=450 [taking . square. roots. of . both . the . sides] [/tex]

u=15√2 m/s

The velocity of car in km/h will be 54√2 or  76.14km/hr [multiplying by 18/5]


HOPE IT Helps!!!!!!!!!!!!!!!!!!!