v = u + at
12 = 0 +(0.8*t)
Time taken to accelerate,t= XX/(0.8) = XX XXXXXXX [t1 = 15 seconds]
XXXXXXXX traveled while accelerating XX XX m/s, s = ut + (X/2)*a*(t^2)
s = (X*t) + X.5* X.8 *(15*15) = XX m
XXXXX 2:XXXXXX proceeds XX XX m/s until XXX XXXXXX XXX applied; it XXXXX to rest at XXXXX X, XX m XXXXXX XXX XXXXX where XXX XXXXXX were applied
XXXXX XXXXXXXX = XXX m
Initial distance XXXXXXX XXXXX XXXXXXXXXXXX = XX m
Distance XXXXXXX XXXXX de-accelerating = 42 m
XXXX, XXXXXXXX covered XX XX m/s uniform speed =XXX - (XX+XX) = XXX - XXX = 168 m
Time XXXXX XX cover XXX m,
t = distance XXXXXXX / uniform XXXXX = 168 / XX =14 XXXXXXX [t2 = 14 seconds]
From XX m/s, it comes XX XXXX (XXXXX XXXXX = 0)
v = u +at
X = XX +XX
XX, XX = -XX
XXXXXXXX to de-accelerate = XX m
Using, s = ut + (X/2)*a*(t^2)
XX = (12*t)+(X/X)*a*(t^t)
XX = t *(12 +(1/2)* a* t)
XXX value XX = -12
XX = t * (XX - X )
t = 7 seconds [XX = X XXXXXXX]
Total time = t1 + XX +t3 = 15 + 14 +7 = 36 XXXXXXX