The following formula relates the distance S (in feet) traveled by a vehicle during an emergency stop to the time T (in seconds) it takes to arrive at a complete stop: s(t)=16t^{2}
At the scene of an accident, Bill uses skid marks to determine the braking distance of one of the vehicles involved was about 650 feet.
What is the corresponding braking time in seconds?
Algebra
650ft/117.3ft/s=5.5 Is NOT the answer. Nor is 40.6 nor 7.4 nor 8.8 nor 16.2 nor 6.5. Nor is 0! Thinking they didn't stop at all before the crash!
To find the corresponding braking time in seconds, we need to solve the given equation for time (T). The formula provided is:
s(t) = 16t^2
Where s(t) is the distance traveled by the vehicle during an emergency stop and t is the time taken to come to a complete stop.
In this case, we are given that the braking distance (s) is 650 feet. So we can rewrite the equation as:
650 = 16t^2
To solve for t, we need to isolate t on one side of the equation.
Dividing both sides by 16, we get:
650/16 = t^2
Simplifying further:
40.625 = t^2
To find t, we take the square root of both sides:
√(40.625) = √(t^2)
This gives us:
t = ±6.375
Since time cannot be negative, we discard the negative solution, leaving us with the positive solution:
t ≈ 6.375 seconds
Therefore, the corresponding braking time in seconds is approximately 6.375 seconds.