Question

A car starts from rest at a stop sign. It accelerates at 2.0 m/s2 for 5.0 seconds, coasts for 3.0 s , and then slows down at a rate of 1.5m/s2 for the next stop sign. Part A How far apart are the stop signs?

Answer 1

88.33 m

Step-by-step explanation:

For AB:

u = 0, t = 5 s, a = 2 m/s^2

Let the distance between A and B be s1 and the velocity at B is v.

Use second equation of motion

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

`s_(1)=0 + 1/2 * 2* 5 * 5`

s1 = 25 m

Use first equation of motion

v = u + a t

v = 0 + 2 x 5 = 10 m/s

For BC:

Let the distance from B to C is s2.

time = 3 s,

speed, v = 10 m/s

s2 = v x t = 10 x 3 = 30 m

For CD:

final velocity, v' = 0

initial velocity, v = 10 m/s

acceleration, a = - 1.5 m/s^2

Let the distance from C to D is s3.

Use third equation of motion

`v'^(2)=v^(2)+2as_(3)`

`0^(2)=10^(2)-2 *1.5 * s_(3)`

s3 = 33.33 m

The total distance covered is s

s = s1 + s2 + s3 = 25 + 30 + 33.33 = 88.33 m

Answer 2

88.3m

Step-by-step explanation:

In this problem there are three sections

you must state the equations as appropriate

attached solution