Question

A car starts from rest at a stop sign. It accelerates at 3.8 m/s 2 m/s2 for 6.0 s, coasts for 1.6 ss , and then slows down at a rate of 3.3 m/s 2 m/s2 for the next stop sign.

How far apart are the stop signs?

Answer 1

To solve this problem we will start by calculating the distance traveled while relating the first acceleration in the given time. From that acceleration we will calculate its final speed with which we will calculate the distance traveled in the second segment. With this speed and the acceleration given, we will proceed to calculate the last leg of its route.

Expression for the first distance is

`s_1 = ut +(1)/(2) at^2`

`s_1 = 0+(1)/(2) (3.8)(6)^2`

`s_1 = 68.4m`

The expression for the final speed is

`v = v_0 +at`

`v = 0+(3.8)(6)`

`v = 22.8m/s`

Then the distance becomes as follows

`s_2 = vt`

`s_2 = (22.8)(1.6)`

`s_2 = 36.48m`

The expression for the distance at last sop is

`v_1^2=v_0^2 +2as_3`

`22.8^2 = 0+2(3.3)s_3`

`s_3 =78.7636m`

Therefore the required distance between the signs is,

`S = s_1+s_2+s_3`

`S = 68.4+36.48+78.76`

`S = 183.64m`

Therefore the total distance between signs is 183.54m