Starting from 130 feet away, a person on bicycle rides towards a checkpoint and then passes it. The rider is traveling at a constant rate of 25 feet per second. The distance between the bicycle and the checkpoint is given by the equation. At what times is the bike 15 feet away from the checkpoint?
(1 point)
Responses
4.6 sec and 9.2 sec
2.9 sec and 5.8 sec
4.6 sec and 5.8 sec
2.9 sec and 3.3 sec
The distance between the bicycle and the checkpoint is given by the equation d = 130 - 25t, where d is the distance and t is the time in seconds.
We need to find the times when the bike is 15 feet away from the checkpoint. So we set d = 15 and solve for t:
15 = 130 - 25t
25t = 130 - 15
25t = 115
t = 115/25
t = 4.6 seconds
Thus, the bike is 15 feet away from the checkpoint at 4.6 seconds.
Now, let's check the other possible solutions:
2.9 seconds:
d = 130 - 25*2.9
d = 130 - 72.5
d ≈ 57.5 feet (not 15 feet)
3.3 seconds:
d = 130 - 25*3.3
d = 130 - 82.5
d ≈ 47.5 feet (not 15 feet)
5.8 seconds:
d = 130 - 25*5.8
d = 130 - 145
d = -15 feet (not a valid distance)
So, the only valid solution is t = 4.6 seconds. The bike is 15 feet away from the checkpoint at 4.6 seconds. The correct option is therefore:
4.6 sec and 9.2 sec.
To find the times when the bike is 15 feet away from the checkpoint, we need to solve the given equation:
Distance = 130 - 25t
We set Distance = 15 and solve for t:
15 = 130 - 25t
Rearranging the equation:
25t = 130 - 15
25t = 115
Dividing both sides by 25:
t = 115/25
t = 4.6
Therefore, the bike is 15 feet away from the checkpoint at 4.6 seconds.
Answer: The bike is 15 feet away from the checkpoint at 4.6 seconds.
Alternatively, we can solve for t by setting the distance equation to 15 and solving:
15 = 130 - 25t
25t = 130 - 15
25t = 115
t = 115/25
t = 4.6
Therefore, the bike is 15 feet away from the checkpoint at 4.6 seconds.
Answer: The bike is 15 feet away from the checkpoint at 4.6 seconds.
In summary, the bike is 15 feet away from the checkpoint at 4.6 seconds.
To find when the bike is 15 feet away from the checkpoint, we need to set up an equation using the given information and then solve it.
Let's start by defining the variables:
- D(t) represents the distance between the bike and the checkpoint at time t.
- t represents the time in seconds.
We are given that the person on the bicycle is traveling at a constant rate of 25 feet per second. This means that the rate is constant and doesn't change.
From the given information, we know that the person starts 130 feet away from the checkpoint. This gives us the initial condition D(0) = 130.
Since the rate is constant, we can write the equation for the distance as:
D(t) = 130 - 25t
The term -25t represents the distance traveled by the bike in t seconds.
To find when the bike is 15 feet away from the checkpoint, we set D(t) equal to 15 and solve for t:
130 - 25t = 15
Simplifying the equation:
-25t = 15 - 130
-25t = -115
Dividing both sides by -25:
t = (-115) / (-25)
t = 4.6
So, the bike is 15 feet away from the checkpoint at 4.6 seconds.
To verify this answer, let's substitute t = 4.6 into the equation for D(t):
D(4.6) = 130 - 25(4.6)
D(4.6) = 130 - 115
D(4.6) = 15
The distance is indeed 15 feet, confirming that t = 4.6 seconds is correct.
Now, we need to check the answer choices to see which other options are correct.
Let's substitute t = 2.9 into the equation for D(t):
D(2.9) = 130 - 25(2.9)
D(2.9) = 130 - 72.5
D(2.9) = 57.5
The distance is not 15 feet, so t = 2.9 seconds is incorrect.
Let's substitute t = 9.2 into the equation for D(t):
D(9.2) = 130 - 25(9.2)
D(9.2) = 130 - 230
D(9.2) = -100
The distance is negative, so t = 9.2 seconds is incorrect.
Let's substitute t = 5.8 into the equation for D(t):
D(5.8) = 130 - 25(5.8)
D(5.8) = 130 - 145
D(5.8) = -15
The distance is negative again, so t = 5.8 seconds is incorrect.
The only correct answer is 4.6 sec and 9.2 sec.