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Question 11
A research engineer is testing the effectiveness of the braking system of a car when it is driven in wet conditions. The engineer measures and records the braking d... show full transcript
Step 1
Answer
Figure 6 shows a linear relationship between log_{10} d and log_{10} V, indicating that as the braking distance d increases, the speed V also increases exponentially. This suggests that d could be modeled as a power of V in the form d = kV^n, where n is a constant. The specific point (0, -1.77) indicates that when log_{10} V is -1.77, the corresponding value of log_{10} d is 0. This provides support for the model since the relationship appears to be consistent across the data points.
Step 2
Answer
From Figure 5, we have a data point (30, 20), which represents V = 30 kmh^-1 and d = 20 m. Substituting this into the model, we get:
20 = 0.017 × 30^n
To isolate n, we first calculate:
20 / 0.017 = 30^n
Calculating the left-hand side yields:
1176.47 = 30^n
Now, take the logarithm base 10:
log_{10}(1176.47) = n × log_{10}(30)
Calculating gives:
n ≈ 2.08, rounded to 3 significant figures, gives n = 2.08.
Step 3
Answer
First, calculate the stopping distance for Sean’s car driving at 60 kmh^-1 (or 60/3.6 m/s):
Stopping distance can be modeled with d = kV^n:
First convert speed: V = 60 kmh^-1 = 16.67 m/s.
Now substitute:
d = 0.017 × (16.67)^2.08
Calculating gives:
d ≈ 4.292 m.
Since the puddle is 100 m away, and Sean takes 0.8 seconds to react, during which he continues moving:
Distance covered during reaction time: 16.67 m/s × 0.8s = 13.336 m.
Thus, total distance until stop = stopping distance + distance during reaction = 13.336 m + 4.292 m = 17.628 m.
This is obviously less than 100 m, therefore Sean will be able to stop before reaching the puddle.
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