Start from 110 feet away, a person on a bicycle rides toward a checkpoint a nd then passes it. The rider is traveling at a constant of 30m feet per second. The distance between the bicycle and the checkpoint is given by the equation d=(110-30t). At what times is the bike 40 feet away from the checkpoint?
My answer is 2.3 seconds for the beginning but I'm not sure if the last number is 4.6 seconds or 5.0 seconds
Well, I guess the bicycle is really taking its time to reach the checkpoint. It's like it's on a leisurely vacation, just enjoying the view along the way.
Let's crunch some numbers here. To find the times when the bike is 40 feet away from the checkpoint, we need to solve the equation d = 40. Here, d represents the distance between the bike and the checkpoint, and t represents the time.
So, let's plug in the given equation d = 110 - 30t into the equation d = 40 and solve for t:
110 - 30t = 40
Now we need to perform some mathematical magic. Let's subtract 110 from both sides:
-30t = -70
And now, let's divide both sides by -30 (or multiply by -1/30, if you want to keep things fancy):
t = 7/3
So, according to my calculations, the bike is 40 feet away from the checkpoint at t = 7/3 seconds. Now, let's convert that to a decimal:
t ≈ 2.33 seconds
So, your initial hunch was right! The bike passes the checkpoint at approximately 2.33 seconds. As for your other options of 4.6 seconds or 5.0 seconds, well, that would've been one really slow bicycle. You could probably walk faster!
But hey, it's always good to double-check these things, so make sure to review the calculations and feel free to ask if you have any more questions. Keep pedaling and keep math-ing!
To find the times at which the bike is 40 feet away from the checkpoint, we need to set the distance equation d = (110-30t) equal to 40 and solve for t.
So, we have the equation: 40 = (110 - 30t)
Let's solve it step-by-step:
Step 1: Subtract 110 from both sides:
40 - 110 = -30t
Step 2: Simplify:
-70 = -30t
Step 3: Divide both sides by -30:
-70 / -30 = t
Step 4: Calculate the quotient:
t ≈ 2.33
Based on the rounded value, we can conclude that the bike is approximately 40 feet away from the checkpoint at around 2.33 seconds.
Now, let's check the other possibility:
Step 5: Substitute t = 2.33 back into the original equation:
d = 110 - 30(2.33)
d ≈ 110 - 69.9
d ≈ 40.1
So, at approximately 2.33 seconds, the bike is around 40.1 feet away from the checkpoint.
Therefore, the correct times at which the bike is 40 feet away from the checkpoint are 2.33 seconds. There is no need to consider 4.6 seconds or 5.0 seconds as they do not satisfy the given equation.
To find the time at which the bike is 40 feet away from the checkpoint, we can set up the equation:
d = 110 - 30t
where d is the distance between the bicycle and the checkpoint, and t is the time in seconds.
We want to find the times when d = 40 feet. So we can substitute d = 40 into the equation:
40 = 110 - 30t
Now, we can solve this equation to find the value of t:
110 - 30t = 40
Subtract 110 from both sides:
-30t = -70
Divide both sides by -30:
t = 70 / 30
t ≈ 2.33 seconds
Therefore, the bike is approximately 40 feet away from the checkpoint at t ≈ 2.33 seconds.
Now, let's check the other option:
We know that the bike passed the checkpoint, and it's moving towards it. So, at some point, the distance between the bike and the checkpoint will be 40 feet. We need to find the time when the distance changes from being greater than 40 feet to being less than 40 feet.
Let's find the time at which the bike reaches the checkpoint:
110 - 30t = 0
-30t = -110
t = 110 / 30
t ≈ 3.67 seconds
Therefore, the bike reaches the checkpoint at t ≈ 3.67 seconds.
Since the bike is moving towards the checkpoint, at t = 2.33 seconds, it would be before reaching the checkpoint, so the last number should be t ≈ 3.67 seconds.
So, your answer is correct. The bike is approximately 40 feet away from the checkpoint at t = 2.33 seconds, and it reaches the checkpoint at t = 3.67 seconds.