challenge problem no.2?

Question

a driver travelling at 70km/hr tries to stop the car and finds that the breaks have failed. the emergency break is then pulled and the car comes to a stop in 120m. find the car's decceleration.


with solution please...thnks

Answer 1

s = 120 m = 0.12 km
u = 70 km/hr
v = 0 km/hr

By third law of Kinematics:

v² = u² + 2*a*s

0 = 70² + 2*a*0.12

0 = 4900 + 0.24*a

a = -4900/0.24 = -20416.66 km/hr²

car's decceleration = 20416.66 km/hr²


OR

for answer in SI unit: m/s²

s = 120 m
u = 70 km/hr = 70*5/18 m/s = 19.44 m/s
v = 0 km/hr

By third law of Kinematics:

v² = u² + 2*a*s

0 = 19.44² + 2*a*120

0 = 378.086 + 240*a

a = -378.086 /240 = -1.5753 m/s²

car's decceleration = 1.5753 m/s²

Answer 2

The two formula's needed are:
V(t) = V(t=0) + at
and
S(t) = V(t=0) t + 0.5 a t^2

where
t = time
a = acceleration
V(t) = velocity
S(t) = distance
V(t=0) = 70 km/h = 19.44 m/s

We obtain:
0 = 19.44 + at
=> a = -19.44/t
and
120 = 19.44 t + 0.5 a t^2

=> 120 = 19.44t - 9.72 t (since a = -19.44/t)
=> 120 = 9.72t
=> t = 12.34 s
=> a = -19.44/12.34 = -1.575 m/s^2