# Assuming a drivers reaction time of 0.25 s, MAKE a table showing stopping distances for a car travelling at initial speeds of 36kph,72kph and 90kph with uniform acceleration of a) 8 m/s^2(in...

Assuming a drivers reaction time of 0.25 s, MAKE a table showing stopping distances for a car travelling at initial speeds of 36kph,72kph and 90kph

with uniform acceleration of

a) 8 m/s^2(in dry concrete)

b) 4 m/s^2(in wet concrete)

-SHOW SOLUTIONS PLEASE. THANKS

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a) You should use the following relation to find the stopping distance `d_s` such that:

`v^2_f = v^2_0 + 2a*d_s`

`v_f` represents the final speed

`v_0 ` represents the initial speed

a represents uniform acceleration

`d_s` represents the stopping distance

Converting the units of measure of initial speed yields:

`36kph = 10m/s ; 72kph = 20 m/s; 90kph = 25 m/s`

Notice that the final speed is 0 such that:

`0 = 10^2 + 2*(-8)*d_s=> d_s = -100/-16 => d_s = 25/4 = 6.25 m`

`0 = 20^2 + 2*(-8)*d_s => d_s = 400/16 = 25 m`

`0 = 25^2 + 2*(-8)*d_s => d_s = 625/16 ~~ 39 m`

**Hence, evaluating the stopping distances, in dry concrete, under the given conditions yields `d_s = 6.25 m, d_s = 25 m , d_s ~~ 39 m.` **

b) You should evaluate the stopping distances iN wet concrete such that:

`0 = 10^2 - 2*4*d_s => d_s = 100/8 = 12.5 m`

`0 = 20^2 - 8d_s => d_s = 400/8 = 50 m`

`0 = 25^2 - 8d_s => d_s= 625/8 = 78.125 m`

**Hence, evaluating the stopping distances, in wet concrete, under the given conditions yields `d_s = 12.5 m, d_s =50 m , d_s = 78.125 m.` **