Solution
x(t) = 8 cos t, x(5π/6)= 8 cos(<span>5π/6)
</span>cos(5π/6)=cos(3π/6 + 2π/6 )=cos(π/3 +π/2)= - sin π/3 (cos (x+<span>π/2)= -sinx)
</span>x(t) = -8sin <span>π/3 = - 4 .sqrt3
</span>v(t) = -8sint = -8sin (π/3 +<span>π/2)= -8 cosπ/3 </span>(sin (x+π/2)= cosx)
v(t) =<span> -8 cosπ/3 = -8/2= - 4
</span>a(5π/6) = - 8cost = -(- sin π/3)= 4 .<span>sqrt3
</span>a(5π/6) = 4 .<span>sqrt3</span>
Your average speed was
(100 m) / (13.8 s) = 7.25 m/s .
If you finished 0.001s ahead of him, then at your average speed, that corresponds to
(7.25 m/s) x (0.001 s) = 0.00725 m
That's 7.25 millimeters ... about 0.28 of an inch !
NOTE:. I think this is only valid if your speed was a constant ~7.25 m/s all the way.
Let the mass of 2500 kg car be 
and it's velocity be 
and the mass of 1500 kg car be 
and it's velocity be 
.
After the bumping the mass be M and it's velocity be V.
By law of conservation of momentum we have
2500 * 5 + 1500 * 1=4000 * V
V = 14000/4000 = 7/2 = 3.5 m/s
So the velocity of the two-car train = 3.5 m/s
Answer:
momentum=mass×velocity
momentum =400kg×20m/s=8000kg.m/s
