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anygoal [31]
3 years ago
11

Sam is observing the velocity of a car at different times. After two hours, the velocity of the car is 50 km/h. After six hours,

the velocity of the car is 54 km/h.
Do the following:
1. Write an equation in two variables in the standard form that can be used to describe the velocity of the car at different times. Show your work and define the variables used.
2. How can you graph the equation obtained in Part A for the first six hours?
Mathematics
1 answer:
kramer3 years ago
4 0
For the acceleration to be uniform(gradient of equation constant), the car must have an existing velocity at time t = 0.
The acceleration (or gradient) = (54 - 50)/4
= 1 km/h
velocity = acceleration x time + initial velocity
50 = 1 x 2 + Vi
Vi = 48 km/h
The equation:
V(t) = t + 48
Where V(t) is the velocity at time t and t is the time.
The equation may be graphed by keeping time on the x axis and velocity on the y axis and starting the graph at the point (0 , 48)
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The value of a certain car decreases by 16% each year. What is the 1⁄2-life of the car?
svet-max [94.6K]

Answer:

The half life of the car is 3.98 years.

Step-by-step explanation:

The value of the car after t years is given by the following equation:

V(t) = V(0)(1-r)^{t}

In which V(0) is the initial value and r is the constant decay rate, as a decimal.

The value of a certain car decreases by 16% each year.

This means that r = 0.16

So

V(t) = V(0)(1-r)^{t}

V(t) = V(0)(1-0.16)^{t}

V(t) = V(0)(0.84)^{t}

What is the 1⁄2-life of the car?

This is t for which V(t) = 0.5V(0). So

V(t) = V(0)(0.84)^{t}

0.5V(0) = V(0)(0.84)^{t}

(0.84)^{t} = 0.5

\log{(0.84)^{t}} = \log{0.5}

t\log{0.84} = \log{0.5}

t = \frac{\log{0.5}}{\log{0.84}}

t = 3.98

The half life of the car is 3.98 years.

7 0
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