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Suppose that a car can accelerate from 30 mph to 50 mph in 4 seconds. Assuming a constant acceleration, find the acceleration (in miles per second squared) of the car and find the distance traveled by the car during the 4 seconds.

4 seconds is $s(4)=234$ feet.

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Integration

Section 1

Antiderivatives

Integrals

Integration Techniques

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So this time they tell us a car is undergoing celebration. So our acceleration is const. And after four seconds, it is accelerated from 30 miles an hour into 50. So do U minus three report. That's just 20/4, which is just five sellers. Acceleration will be increasing by five MPH every second. Let's see. So they want to know how much the car has traveled during those three seconds. Well, we know that our velocity, let's see, is equal. Teoh starting at 30 and adding five every additional second. So we're gonna need to take the position function, and we can take the position at the last second. Subtract the position that sex Teoh what we want. Well, let's see the anti derivative. This is fine. He squared over two plus 30 plus some another. See? Okay, whenever we want to know the difference in positions as a sorry 19 seconds. So when we do s for attract s zero will be subtracting this unknown constant from itself. So it's gonna cancel out. It's irrelevant to us here, So let's just plug in for and we'll get let's see. Five time 16 over to laws 30 times for the track bull zero plus zero. So let's complete this. This isn't AIDS, so we have 40 plus 1 20 1 60 You can assume that they've traveled the 1 60 miles.

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