Acceleration Deceleration Physics Problem

[spaceshipone]

A car has a maximum constant acceleration of 10 ft/s^2 and a maximum constant deceleration of 15 ft/s^2. determine the minimum amount of time it would take to drive one mile assuming the car starts and ends at rest and never exceeds the speed limit 55 mi/hr.

Why do you need to know the deceleration constant if we are talking about the car accelerating and going to 55 mi/hr?
I would assume you just plug it into the constant acceleration formula and that is it.
I was also thinking maybe it has to accelerate to first at a constant rate and then decelerate to slow to 55 mph. I which case I have no idea how to solve the problem.

Anyone want to take a stab at this?

 

[rl.bhat]

Imagine you are driving from place A to place B. What do you do?
You start from rest, then accelerate until you reach the speed limit. You drive some time, then when you are near B, you apply the brakes to stop the car at B. Total distance is 1 mile.
Find t and S during acceleration and retardation, using appropriate kinematic equation.
Find the remaining distance which covered with uniform velocity. From that find the time taken to cover that distance. So you can find the total time to cover the whole distance.

 

[tomcue]

I would like to clarify that acceleration and deceleration are two different concepts in physics. Acceleration refers to the rate of change of velocity, while deceleration refers to the rate of change of negative velocity. In this problem, we are dealing with both acceleration and deceleration, as the car needs to accelerate to reach the speed limit and then decelerate to maintain it.

To solve this problem, we can use the basic equation of motion, s = ut + 1/2at^2, where s is the distance traveled, u is the initial velocity (which is zero in this case), a is the acceleration, and t is the time taken. We can use this equation for both the acceleration and deceleration phases of the car's motion.

First, we need to determine the time it takes for the car to accelerate from rest to 55 mi/hr. The maximum acceleration of the car is given as 10 ft/s^2, which can be converted to 0.2 mi/hr^2. Using this value for a, and s = 1 mile (which is equivalent to 5280 ft), we can rearrange the equation to solve for t. This gives us t = √(2s/a) = √(2*5280/0.2) = √52800 = 229.99 seconds.

Next, we need to determine the time it takes for the car to decelerate from 55 mi/hr to rest. The maximum deceleration of the car is given as 15 ft/s^2, which can be converted to 0.3 mi/hr^2. Again, using this value for a, and s = 1 mile, we can solve for t to get t = √(2s/a) = √(2*5280/0.3) = √35200 = 187.08 seconds.

Finally, we can add these two times to get the total time taken by the car to travel one mile, which is 229.99 + 187.08 = 417.07 seconds or approximately 6.95 minutes.

In conclusion, the deceleration constant is necessary to accurately determine the total time taken by the car to travel one mile, as the car needs to accelerate and then decelerate to maintain a constant speed of 55 mi/hr. I hope this explanation helps in understanding the problem