To stop a car, you require first a certain reaction time to begin braking. Then the car slows under the constant braking deceleration. Suppose that the total distance moved by your car during these two phases is 56.7 m when its initial speed is 82.5 km/h, and 24.4 when the initial speed is 50.3 km/h. What is your reaction time? What is the magnitude of the deceleration?
Let tr be reaction time.
The distance traveled during the reaction time is vi*tr, so the distance traveled during stopping is (distancetotal-vi*tr). Now you have an equation...
Vf^2=Vi^2 + 2ad Use that. I will set up one equation for you, you do the second, and solve for tr, and a.
1) 0=24.4^2 + 2a (56.7-24.4tr)
2) ....
I still don't get how to get the time it takes to get from 82.5 to 50.3 km/h, i know its a constant deceleration, but i don't know how to get the time
I still don't get how to get the time it takes to get from 82.5 to 50.3 km/h, i know its a constant deceleration, but i don't know how to get the time it takes to decelerate
I still don't get how to get the time it takes to decelerate from 82.5 to 50.3 km/h, i know its a constant deceleration, but i know how to find either the deceleration or the time
To find the reaction time and the magnitude of the deceleration, we can use the equations of motion relating distance, initial speed, acceleration, and time.
Let's break down the problem step by step:
Step 1: Convert the initial speeds from km/h to m/s. Since the units of acceleration are in m/s^2, it's easier to work with consistent units.
82.5 km/h = (82.5 × 1000) / (60 × 60) = 22.92 m/s
50.3 km/h = (50.3 × 1000) / (60 × 60) = 13.97 m/s
Step 2: Calculate the distance covered during the reaction time.
The distance covered during the reaction time can be represented as:
D1 = (1/2) * a * t1^2
Where D1 is the distance covered, a is the deceleration, and t1 is the reaction time.
Step 3: Calculate the distance covered during the constant braking deceleration.
The distance covered during the constant braking deceleration can be represented as:
D2 = v * t2 - (1/2) * a * t2^2
Where D2 is the distance covered, v is the initial speed, a is the deceleration, and t2 is the time during the constant braking deceleration.
Step 4: Use the given data to form equations.
From the problem statement, we have two equations:
Equation 1: D1 + D2 = 56.7 m
Substituting the expressions from steps 2 and 3, we have:
(1/2) * a * t1^2 + (v * t2 - (1/2) * a * t2^2) = 56.7
Equation 2: D1 + D2 = 24.4 m
Again, substituting the expressions from steps 2 and 3, we have:
(1/2) * a * t1^2 + (v * t2 - (1/2) * a * t2^2) = 24.4
Step 5: Solve the equations simultaneously to find the unknowns.
By solving the simultaneous equations, we can find the values of t1 (reaction time) and a (deceleration).
This requires all the variables to be known except t1 and a, which we can substitute from the given data:
a * t1^2 + (v * t2 - (1/2) * a * t2^2) = 56.7
a * t1^2 + (v * t2 - (1/2) * a * t2^2) = 24.4
Simplifying the equations further:
t1^2 * a - t2^2 * a * 0.5 = 56.7 - v * t2
t1^2 * a - t2^2 * a * 0.5 = 24.4 - v * t2
Now we have two equations with two unknowns (t1^2 * a and t2^2 * a). We can solve these equations to find the values of t1 and a.
Using the given data, we have:
56.7 - (22.92 * t2) = t1^2 * a
24.4 - (13.97 * t2) = t1^2 * a
Solving these equations will give us the values of t1 (reaction time) and a (deceleration).