solve the problem..please?

Question

albert is riding his scooter at a velocity of 80 km/hour, when he sees an old woman crossing the road 45m away. he immediately steps hard on the breaks to get the maximum decceleration of 7.5 m/s^2. how far will he go before stopping? will he hit the old woman or not?



please answer...with solutions..

Answer 1

time to stop t(s)
use v^2(before)-v^2(after)=2*a*s
=> 200/9^2(m/s)-0=2*7.5*s

=>s=32.92<45

so he won't hit the old woman

Answer 2

80km/hr = 80,000m/hr = 80,000/360 m/s = 200/9 m/s

u (starting velocity) = 200/9 m/s
v (final velocity) = 0 m/s
a (acceleration) = -7.5m/s^2
Trying to find s (distance traveled)

v^2=u^2+2as
Solve for s
s=(v^2-u^2)/(2a)
s=(0^2-(200/9)^2)/(2*-7.5)
s=(-(200/9)^2)/-15
s=(-40,000/81)/-15
s=40,000/1,215
s=32.92m

32.92<45
Albert will not hit the old woman

Answer 3

WE KNOW THAT
V(sq)-U(sq)=2aS
WHERE V-FINAL VELOCITY,U-INITIAL VELOCITY,A-DECELERATION,S-DISTANCE.
V=0;U=80KM\H=22.22M\S;A=7.5 SUBSTITUTING WE GET S=32.92M
SO IT WILL NOT HIT THE WOMEN