The distance traveled by an object moving in a straight line is given by s = t2 − 2t,
where s is in feet and t is the time in seconds the object has been in motion. How long
(to the nearest tenth) will it take the object to move 48 feet?
To find the time it will take for the object to move 48 feet, we can substitute the given distance (48 feet) into the equation and solve for t.
The given equation is s = t^2 - 2t, where s represents the distance in feet and t represents the time in seconds.
Substituting s = 48 feet into the equation, we get:
48 = t^2 - 2t
Rearranging the equation:
t^2 - 2t - 48 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula. Let's use the quadratic formula.
The quadratic formula is given by:
t = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = 1, b = -2, and c = -48.
Substituting these values into the quadratic formula:
t = (-(-2) ± √((-2)^2 - 4(1)(-48))) / (2(1))
t = (2 ± √(4 + 192)) / 2
t = (2 ± √196) / 2
t = (2 ± 14) / 2
Now we have two possible solutions:
1. t = (2 + 14) / 2
2. t = (2 - 14) / 2
Simplifying each equation:
1. t = 16 / 2 = 8
2. t = -12 / 2 = -6
Since time cannot be negative in this context, we discard the second solution. Therefore, the only valid solution is t = 8.
Hence, it would take approximately 8 seconds for the object to move 48 feet.
just solve
t^2-2t = 48
t^2-2t-48 = 0
(t-8)(t+6) = 0
I expect you can take it from there, no?