Motion
Q. A bus starts from rest at point A and accelerates at the rate of 0.8 m/s^2 unit it reaches a speed of 12 m/s. Its then proceeds at XX m/s until XXX XXXXXX XXX applied; it comes to XXXX at XXXXX X, XX m XXXXXX XXX XXXXX XXXXX XXX XXXXXX were applied. XXXXXXXX uniform XXXXXXXXXXXX XXX XXXXXXX that XXX distance XXXXXXX X and B is XXX m, XXXXXXXXX XXX XXXX XXXXXXXX XXX the bus XX XXXXXX from X to X.
XXX.XXX question has 3 XXXXX. XXX first XXXX XX XXXXXXXXXXX motion, then XXX uniform XXXXXX and XXXX XXXX is retardation XXXXXX.
Let the XXXXXXXXXXX XXXX XX XX, the uniform XXXXXX XXXX XX DE XXX retardation part is XX.
The accelerated XXXX AD :
TimetX= (12 – X) / X.8 =15 s
Distance AD = 0*XX + X*0.8*XX2 = XX m
The XXXXXXX motion XXXX DE :
Distance XX = AB – (AD + XX) = XXX m - (90 + XX) = XXX m
TimetX= 168 m / 12 m/s =XX s
XXX XXXXXXXXXXX part XX :
XXXX tX= (XX – 0) / aX = XX/aX
Or a = XX/ t3
XXXXXXXX EB = XX m
XXXXX, XX* tX - X*a3* t32 = XX* t3 - X*XX/ t3* tXX
= 12* t3 - XXX = 6t3
So, XX3= XX ORtX= X s
a3 = XX/X m/sX
XXXX,total XXXX required for XXX XXX to XXXXXX XXXX A to B
= t1+ tX+ tX = XXX +14s + XX =36 s
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