So, how far is the car from where it was at t = 0 is 40 m
<h3>Velocity of the car</h3>
Since the location x of the car in meters is given by the function x = 30t - 5t² where t is in seconds, we need to find the time at which its velocity is 10 m/s in the negative direction by differentiating x with respect to t to find its velocity, v.
So, v = dx/dt
= d(30t - 5t²)/dt
= d30t/dt - d5t²/dt
= 30 - 10t
When v is 10 m/s in the negative direction, v = -10 m/s.
So, v = 30 - 10t
-10 = 30 - 10t
-10 - 30 = -10t
-40 = -10t
t = -40/-10
t = 4 s
<h3>The distance at 4 s when its velocity is -10 m/s</h3>
Since at t = 4 s, its velocity is -10 m/s and x = 30t - 5t² is the car's location. The car's distance from t = 0 after its velocity is -10 m/s is
x(4) - x(0) = 30(4) - 5(4)² - [30(0) - 5(0)²]
= 120 - 5(16) - [0 - 0]
= 120 - 80 - 0
= 40 m
So, how far is the car from where it was at t = 0 is 40 m
Learn more about distance of car here:
brainly.com/question/17097458
You would have a maximum of 6 days
$160-$10 = $150
$150/$25=$6
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Step-by-step explanation:
I wonder if it wants 3.80 per gallon?
Perhaps the imperial value which would have been 4.50 per gallon which is an imperial gallon (used in Canada). It's hard to tell. I would have thought 3.78 should be more than close enough.
Answer:
Step-by-step explanation:
hope it helps..
have a great day!!
