Question

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?

Answer 1

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,
`v_(o) = 5 m/s`

The minimum value of acceleration ,
`a_(min) = - 1 m/s^(2)`

The maximum value of acceleration ,
`a_(max) = 1 m/s^(2)`

Now,

When the deceleration starts the ferry slows down and at minimum deceleration of
`- 1 m/s^(2)`, 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:

`v'^(2) = v_(o)^(2) + 2ad`

`0^(2) = 5^(2) + 2* (- 1)d`

`d = 12.5 m`