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Sagot :
Answer:
A. 2.3 sec and 5 sec
Step-by-step explanation:
We are given that start from 110 feet away, a person on a bicycle rides towards a checkpoint and then it passes it. the rider is traveling at a constant rate of 30 feet per second.
The distance between the bicycle and the checkpoint is given by the equation :
[tex]d=|110-30t|[/tex]
Now we are required to calculate at what times is the bike 40 feet away from the checkpoint. So, substitute d = 40 in the equation
[tex]40=|110-30t|[/tex]
[tex]40=110-30t[/tex] and [tex]40= -110+30t[/tex]
[tex]30t=110-40t[/tex] and [tex]40+110=30t[/tex]
[tex]30t=70[/tex] and [tex]150=30t[/tex]
[tex]t=\frac{70}{30}[/tex] and [tex]\frac{150}{30}=t[/tex]
[tex]t=2.3[/tex] and [tex]5=t[/tex]
Hence at 2.3 sec and 5 sec the bike is 40 feet away from the checkpoint.
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