A driver sees a bear in his headlights. He is driving at 25m/s and decelerates at a rate of 4.9m/s2.

After breaking for 25.0 m, what is the ranger's speed?

### 2 Answers | Add Yours

We can use equations of motion here.

`v^2 = u^2+2aS`

The data we have is:

`u = 25m/s`

`a = -4.9m/s^2` since we have a deceleration

`S = 25m`

`v^2 = 25^2-2xx4.9xx25`

`v = 19.49m/s`

*So the speed of ranger after traveling 25m breaking distance is 19.49m/s*

**Sources:**

V^2- U^2 = 2as

Now, v=unknown

U= 25m/s

a=-4.9m/s^2

s= 25m

Solving we get,

V^2= -2*4.9*25+25*25=625-245=380

V= √380=19.49m/s

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes