Grade 12th passMechanics
Question 14 with explanation and without graphical method
Question 14 with explanation and without graphical method
1 Answer
Arun
Dear Yash
3 stages of motion: acceleration, constant velocity, deceleration
deceleration:
vf=vi+at
0 m/s = 24 m/s + (-4 m/s2)t
t=6 s
Δx =vi*t + 1/2* a* t2
Δx=(24m/s)(6s)+1/2 *(-4 m/s2)(6s)2
Δx=72 m
acceleration and constant velocity
time for acceleration t1 and constant velocity t2
t1+t2=50
distance for acceleration d1 and constant velocity d2
d1+d2=960
d1 = (vi+vf)/2)*t1
d1 = (0+24)/2)*t1
d1 = 12 m/s * t1
d2 = 24 m/s * t2
960 = 12 m/s * t1 + 24 m/s * t2
960 = 12 m/s *(50-t2) + 24 m/s * (t2)
960 = 12 m/s *t2 + 1200
360 = 12 m/s *t2
t2=30 s
t1=20 s
solve for a
vf = vi +at
24 m/s = 0 m/s + a (20 s)
a = 1.2 m/s2
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