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Physics 19 Online
OpenStudy (anonymous):

When you see a traffic light turn red, you apply the brakes until you come to a stop. Your initial speed was 12m/s . Assume constant deceleration. Suppose the car comes to rest in 36m . How much times does this take?

OpenStudy (anonymous):

So you know initial velocity, final velocity, distance, and that it is uniformly accelerated motion. Do you have a formula that includes the given information along with the unknown, delta t?

OpenStudy (anonymous):

no i do not

OpenStudy (anonymous):

Alright, perhaps we need an intermediate step. Do you have enough information to calculate the acceleration?

OpenStudy (anonymous):

you can solve the problem using the equation\[\frac{ x }{t }=\frac{v+v_0}{2}\]

OpenStudy (radar):

Using the equation provided by alothman, I got a t of 6 seconds, is that reasonable for a speed of 43.2 km/hr and good brakes?

OpenStudy (danjs):

Given:\[v _{f} = 0~m/s\]\[v _{i}=12~m/s\]\[\Delta x = 36~m\]\[a = constant\] Find: t [s]

OpenStudy (danjs):

Using the kinematic equations with constant acceleration, First calculate the constant acceleration from: \[v _{f}^2=v _{0}^2+2a( \Delta x)\] Use that value for acceleration, and calculate the time t [s] to stop the car \[v _{f}=v _{o} + a*t\]

OpenStudy (radar):

Using the equations provided by DanJS, results in a value of 6 seconds for t. For that speed which is 43.2 km/hr that seems reasonable.

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