If a car is travelling at 60. miles per hour and the brakes are applied so that it accelerates at -15.0 ft/s2, how many feet will it take to stop?
i realize i am posting a lot of questions, but i don't understand them!
First convert 60 mph to 88 ft/s.
During braking, the average velocity will be 1/2 * 88 = 44 ft/s
The time required to stop will be
(88 ft/s)/15 ft/s^2 = 5.87 s
Stopping distance = average speed x time
that tells me the time, but how do i find the number of feet?
No worries! I'm here to help you. Let's break down the problem step by step.
To determine how many feet it will take for the car to stop, we need to find the distance traveled while decelerating.
First, let's convert the car's speed from miles per hour to feet per second, as the acceleration is given in feet per second squared.
1 mile = 5280 feet
1 hour = 3600 seconds
So, 60 miles per hour can be converted as follows:
60 miles/hour * 5280 feet/mile * 1 hour/3600 seconds = 88 feet/second
Now, we have the initial velocity (v₀) of the car as 88 ft/s, and the acceleration (a) as -15.0 ft/s² (negative sign indicating deceleration). We need to find the stopping distance (d).
We'll use the following formula of motion to solve for d:
v² = v₀² + 2ad
Here, v is the final velocity, which will be 0 since the car comes to a stop.
So, substituting the values into the formula:
0² = (88 ft/s)² + 2(-15.0 ft/s²)d
0 = 7744 ft²/s² - 30.0 ft/s²d
30.0 ft/s²d = 7744 ft²/s²
Dividing both sides by 30.0 ft/s² to solve for d:
d = 7744 ft²/s² / 30.0 ft/s²
d ≈ 258.133 feet
Therefore, the car will take approximately 258.133 feet to stop.
I hope this explanation was helpful to you. If you have any further questions, feel free to ask!