need help. The skier goes down the slope distance d and time t. The relationship between d and t can be represented in the following form
d = at = bt^2
where a nd b are constants.
If given d=11 t=2 and d=28 t=4 how to calculate values of a and b. Calculate the length of the slope corect to the nearest metre, if the skier riches the bottom in 11 seconds.
To calculate the values of constants a and b, we can use the given data points where d = 11 when t = 2 and d = 28 when t = 4.
Using the equation:
d = at = bt^2
1. Plug in the values for the first data point (d = 11, t = 2):
11 = 2a = 4b
2. Plug in the values for the second data point (d = 28, t = 4):
28 = 4a = 16b
Now we have two equations with two variables (a and b) that we can solve simultaneously.
From equation 1:
2a = 11/2
Simplifying, we get:
a = 11/4
From equation 2:
4a = 28/16
Simplifying, we get:
a = 7/4
Now, substitute the value of a back into either equation to solve for b:
Using equation 1:
2(11/4) = 4b
Simplifying:
11/2 = 4b
Dividing by 4, we get:
b = 11/8
Now we have found the values of a and b.
To calculate the length of the slope, we substitute the given time of 11 seconds into the equation, with the values of a and b we just found.
d = at = bt^2
Substituting the values:
d = (11/4)(11) = (11/8)(11^2)
Simplifying, we get:
d = 121/4 = 133.25 meters (rounded to the nearest meter).
Therefore, the length of the slope, to the nearest meter, is 133 meters.