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never [62]
2 years ago
7

Greg and Marti are racing remote control cars. The race starts at t=0 and the car’s distance is measured from the start line. Bo

th cars have a constant rate of change. Information about each of the cars during the race is shown below.
The race was over after 10 seconds. The winner of the race is defined by the person who is farthest from the start line. Whose car should be crowned the winner? Use mathematics to justify your answer.

Mathematics
1 answer:
Sphinxa [80]2 years ago
4 0

Answer:

Marti

Step-by-step explanation:

Marti is farthest because it says that greg's car travels 13 feet but after two seconds he was at 30.7 away from the start line and marti was 2 feet in front but martyi goes faster because her car goes way more faster every second: (16.5 > 13) so: Greg is 30.7 + 13 x 10 = 160.7 | Marti: 2 + 16.5 x 10 = 167, So Marti wins the race and should be crowned the winner.

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Answer:

Step-by-step explanation:

The x intercept is the value of x when y = 0.  Use the equation to find the value of x:

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A grasshopper jumps straight up from the ground with an initial vertical velocity of 8 feet per second. After how many seconds w
vlabodo [156]

Answer:

Problem 1 : After 0.25 s

Problem 2 : The squirrel doesn't reach the ground before the nut

Step-by-step explanation:

Problem 1

This is a standard physics problem

The equation that defines the movement of the grasshopper given the initial velocity, and the acceleration is the following

Distance =  Do + Vo*t + 0.5*a*t^2

Where

Do is the initial distance.

(Do = 0, because the grasshopper is on the ground and we choose the reference system there)

Vo is the initial velocity = 8 feet/sec

a is the acceleration that is exerted on the body (in this case a = g = -9.8 m/s^2 = - 32.174 feet per second per second.)

t is the time

We substitute in the equation, and solve for t (time)

Distance =  Do + Vo*t + 0.5*a*t^2

(1 foot)=  (0)+ (8 feet/s)*t + 0.5*(-32.174 feet/s^2)*t^2

Solving this equation, we get that

t = 0.25 s

Problem 2

This problem is similar to the previous one

We apply the same formula

D =  Do + Vo*t + 0.5*a*t^2

But in this case, we are going in the opposite direction, and we have an initial distance.

Do = 27 feet

Vo is the initial velocity of the nut = -6 feet/s

D = 0  (Ground)

We substitute in the equation, and solve for t (time)

(0) =  (27) + (-6 feet/s)*t + 0.5*(-32 feet/s^2)*t^2

t = 1.125 s  <  2s

The nut reaches ground in 1.125 s, so the squirrel doesn't reach the ground before the nut

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Step-by-step explanation:

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