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The driver of a car traveling at 54ft/sec suddenly applies the brakes. The position of the car is s=54t-3t^2, t seconds afyer the driver applies the brakes.
How many seconds after the driver applies the brakes does the car come to a stop

Respuesta :

luistaloncarret
  s = 60t - 3t^2 
v = ds/dt = 60 - 6t 

when it comes to rest, v = 0, 
so t = 10 sec 

now distance travelled = time taken x average velocity 
= 10 x (60 + 0)/2 
= 300 ft 

After time  [tex]t = 9[/tex] seconds the driver applies the brakes does the car come to a stop.

What is time?

" Time is defined as the measurable slot of period in which required action is done."

Formula used

[tex]y = x^{n} \\\\\implies \frac{dy}{dx} = nx^{n-1}[/tex]

According to the question,

Position of the car [tex]'s' = 54t - 3t^{2}[/tex]

[tex]'s'[/tex] represents the distance

[tex]'t'[/tex] is the time in seconds

When driver applies break [tex]v= 0[/tex],

[tex]v = \frac{ds}{dt}[/tex]

Calculate the first derivative with respect to time we get,

[tex]'s' = 54t - 3t^{2}\\\\\implies \frac{ds}{dt} = 54- 6t[/tex]

As [tex]\frac{ds}{dt} =0[/tex] we get the required time as per given condition,

[tex]54-6t =0\\\\\implies 6t =54\\\\\implies t = 9[/tex]

Hence, after time  [tex]t = 9[/tex] seconds the driver applies the brakes does the car come to a stop.

Learn more about time here

https://brainly.com/question/15356513

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