v = u + at
12 = 0 +(0.8*t)
Time taken to accelerate,t= XX/(0.X) = 15 seconds [t1 = XX seconds]
XXXXXXXX traveled while accelerating to 12 m/s, s = ut + (X/X)*a*(t^X)
s = (X*t) + X.X* 0.8 *(XX*15) = 90 m
XXXXX 2:Itthen proceeds XX 12 m/s XXXXX the breaks are applied; it comes to rest XX XXXXX X, XX m beyond XXX XXXXX where the breaks were XXXXXXX
XXXXX Distance = XXX m
XXXXXXX XXXXXXXX XXXXXXX XXXXX accelerating = XX m
XXXXXXXX covered XXXXX de-accelerating = 42 m
XXXX, distance XXXXXXX XX 12 m/s uniform XXXXX =300 - (XX+42) = 300 - 132 = 168 m
XXXX taken XX XXXXX XXX m,
t = XXXXXXXX covered / XXXXXXX XXXXX = XXX / 12 =XX seconds [t2 = 14 XXXXXXX]
From 12 m/s, it XXXXX XX XXXX (XXXXX XXXXX = X)
v = u +at
0 = 12 +XX
So, XX = -12
XXXXXXXX XX de-XXXXXXXXXX = XX m
XXXXX, s = ut + (1/2)*a*(t^2)
42 = (12*t)+(X/2)*a*(t^t)
XX = t *(XX +(X/X)* a* t)
XXX XXXXX at = -12
XX = t * (12 - X )
t = 7 seconds [XX = 7 seconds]
Total time = t1 + t2 +XX = XX + XX +7 = 36 XXXXXXX