Answered

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A driver who is moving at 15 m/s, sees a stop light and puts on the brakes 30 m before an intersection.

a) What acceleration will the driver have to have to stop before entering the intersection?

b) How much time will it take to stop the car?

Sagot :

leena

Hi there!

We can use the kinematic equation:

[tex]vf^2 = vi^2 + 2ad[/tex]

vf = final velocity (0 m/s)

vi = initial velocity (15 m/s)

a = acceleration due to breaking (? m/s²)

d = displacement (30 m)

We can plug in the given values:

[tex]0^2 = 15^2 + 2(a)(30)\\\\0 = 225 + 60a[/tex]

Solve:

[tex]-60a = 225 \\\\a = \boxed{-3.75 m/s^2}[/tex]

To solve for the time taken, we can use the following:

[tex]t= \frac{v_f-v_i}{a}[/tex]

Plug in the values:

[tex]t = \frac{0-15}{-3.75} = \boxed{4 sec}[/tex]

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