The baltimore harbor water taxi is approaching the dock with a velocity of v0 = 5 m/s. the water taxi acceleration is limited to −1 m/s < a < 1 m/s, how far from the dock must the ferry begin slowing down if it is to avoid a collision?

In order to avoid collision, the ferry must stop at a distance of 12.5 m from the dock.

Solution:

The initial velocity of the taxi,

The minimum value of acceleration ,

The maximum value of acceleration ,

Now,

When the deceleration starts the ferry slows down and at minimum deceleration of , the ferry stops.

Thus, inthis case, the final velocity, v' is 0.

Now, to calculate the distance covered, 'd' in decelerated motion is given by the third eqn of motion: