Question

at = 0.02 m/s^2 where t is the time in seconds measured from rest. Imagine starting @ C. Calculate the velocity and aceleration at point B.

120 250 m

Answer 1

Now,

Distance to be traveled on straight track, d = 250 m

Now,

radius of the curvature, r = 50 m

From figure we can see angle covered is 120^{$\dpi{100} ^{\circ}$}

Thus, distance to be traveled on curved path, d' = (120/360)*2*$\dpi{100} \pi$*r = 1/3*2*3.14*50

                                                                              = 104.66 m

Thus, total distance traveled, x = d + d' = 250 + 104.66

                                                   = 354.66 m

As we have,

=> x = 0.0033 * t³

=> 354.66 = 0.0033 * t³

=> t³ = 354.66 / 0.0033 = 107472.7

=> t = $\dpi{80} \sqrt[3]{107472.7}$ = 47.54 s

Now,

v = 0.01 * t² = 0.01 * 47.54²

   = 22.6 m/s

and

a = 0.02*t = 0.02 * 47.54

   = 0.95 m/s²