The correct option is B 59.66 ms−1
Given,
distance covered, s = 250 m,
time taken, t = 6 s
acceleration, a = 6 ms−2
Let the initial velocity of the car be u and final velocity be v.
From the second equation of motion,
s=ut+12×at2
250=u×6+12×6×62
⇒ u=23.66 ms−1
From the first equation of motion,
v = u + at
⇒v=23.66+6×6 = 59.66 ms−1
∴The final velocity of the car is 59.66 ms−1