A car is traveling at 104 km/h when the driver sees an accident 50 m ahead and slams on the brakes. What minimum constant deceleration is required to stop the car in time to avoid a pileup

a =  - 8.34 m/sec²       ( deceleration or negative)  

Equations for UAM  ( uniformly accelerated motion) are:

vf = v₀ ±  a*t           and      s  =  s₀  + v₀*t + (1/2)*a*t²

In our case, the motion is with deceleration, then

vf = v₀ - a*t       and    s = s₀  +  v₀*t  - (1/2)*a*t²  

working on these equatios we get:

vf = v₀ - a*t     (1)           s  -  s₀   =  v₀*t  - (1/2)*a*t²      (2)

v₀ - vf = a*t

t  =  (v₀ - vf)/a

By substitution of (1) in equation (2)

s  -  s₀   =  v₀ * (v₀ - vf)/a  -  (1/2) * a* [(v₀ - vf)/a]²

s  -  s₀   =  (v₀² - v₀*vf)/a  -   (1/2) * a* (1/a²)* (v₀ - vf)²

s  -  s₀   =  1/a * ( v₀² - v₀*vf ) - 1/a* (1/2)  * (v₀ - vf)²

s  -  s₀   =  1/a* [  ( v₀² - v₀*vf ) - (1/2) * (v₀ - vf)²]

a * (s  -  s₀ )  =   v₀² - v₀*vf - v₀²/2 - vf²/2 + v₀*vf

a * (s  -  s₀ )  =  (1/2) * v₀² - (1/2)*vf²

a * (s  -  s₀ )  =  (1/2) * ( v₀² - vf²)

We find an expression to calculate the minimum deceleration to stop the car in time to avoid crashing

s₀ = 50 meters            s  =  0        v₀ =  104 Km/h    vf = 0

1 Km  = 1000 m    and   1 h = 3600 sec

v₀ = 104 Km/h    =  28.88 m/sec

a  =  (1/2) [ (28.88)² - 0 ] / 0 - 50

a =  - 8.34 m/sec²       ( deceleration or negative)  

i had this question and the answer was 3.15 miles.

martina travels from school to her mom's office. from there, she walks to school. after school, she walks to the library and then to a friends house before finally ending back at home. find martina's total distance traveled.

a) 0 miles  

b) 0.4 miles  

c) 1 mile  

d) 3.15 miles