Asked by poppy
Three cars start traveling at the same time. The distance traveled by each is represented in three different ways, with y representing miles traveled and t representing time passed in hours.
Use the images to answer the question.
Car 1:
y=t
Car 2:
Distance Traveled by Car 2 at Different Time Intervals
0
0.2
0.4
0.6
0.8
1
0
0.04
0.16
0.36
0.64
1
Car 3:
A coordinate plane goes from negative 5 to 5 on the t-axis in intervals of 1 and from negative 5 to 5 on the y-axis in intervals of 1.
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A graph of the equation y equals 2 to the t power negative1 is shown. It includes the points left parenthesis 0 comma 0 right parenthesis, left parenthesis 1 comma 1 right parenthesis, left parenthesis 2 comma 3 right parenthesis.
Compare the speeds of the 3 cars after 5 hours. In your answer, describe the function represented by each car.
(2 points)
All Answers 1
Answered by
GPT-5 mini
AI
Car 1: y = t. Linear function — constant speed 1 mi/hr. After 5 hr the car has gone 5 mi and its speed is 1 mph.
Car 2: y = t^2 (table matches t2). Quadratic — speed increases linearly: v = dy/dt = 2t. After 5 hr the car has gone 25 mi and its instantaneous speed is v = 2(5) = 10 mph.
Car 3: y = 2^t − 1 (points (0,0),(1,1),(2,3) match). Exponential — speed grows exponentially: v = dy/dt = (ln 2)·2^t. After 5 hr the car has gone 2^5 −1 = 31 mi and its instantaneous speed is (ln 2)·32 ≈ 22.18 mph.
So after 5 hours Car 3 is fastest, then Car 2, then Car 1.
Car 2: y = t^2 (table matches t2). Quadratic — speed increases linearly: v = dy/dt = 2t. After 5 hr the car has gone 25 mi and its instantaneous speed is v = 2(5) = 10 mph.
Car 3: y = 2^t − 1 (points (0,0),(1,1),(2,3) match). Exponential — speed grows exponentially: v = dy/dt = (ln 2)·2^t. After 5 hr the car has gone 2^5 −1 = 31 mi and its instantaneous speed is (ln 2)·32 ≈ 22.18 mph.
So after 5 hours Car 3 is fastest, then Car 2, then Car 1.
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