A car traveling at 42 ft/sec decelerates at a constant 7 feet per second squared. How many feet does the car travel before coming to a complete stop?
Well, if a car is decelerating at a constant rate, it means it's slowing down. So it's going from 42 feet per second to zero. To find out how far it travels, we need to use a little physics. The formula we can use is:
vf^2 = vi^2 + 2ad
where vf is the final velocity (which is zero in this case), vi is the initial velocity (42 ft/sec), a is the acceleration (-7 ft/sec^2), and d is the distance.
Rearranging the formula, we get:
d = (vf^2 - vi^2) / (2a)
Plugging in the values, we have:
d = (0^2 - 42^2) / (2 * -7)
Now, let's simplify:
d = (-1764) / (-14)
Guess what? Two negatives cancel each other out, so we have:
d = 126 feet
So the car travels 126 feet before coming to a complete stop. That's quite a distance! But hey, at least it gives the pedestrians ample time to run away!
To find the distance the car travels before coming to a complete stop, we can use the formula:
v^2 = u^2 + 2as
Where:
- v is the final velocity (0 ft/sec since the car comes to a complete stop)
- u is the initial velocity (42 ft/sec)
- a is the acceleration (-7 ft/sec^2, since it is decelerating)
- s is the distance traveled
Plugging in the given values, we have:
0^2 = 42^2 + 2(-7)s
Simplifying the equation, we get:
0 = 1764 - 14s
Rearranging the equation, we have:
14s = 1764
Dividing both sides of the equation by 14, we get:
s = 126
Therefore, the car travels 126 feet before coming to a complete stop.
To find the distance traveled before the car comes to a complete stop, we first need to determine the time it takes for the car to stop. We can use the formula:
Final velocity (vf) = Initial velocity (vi) + Acceleration (a) * Time (t)
In this case, the final velocity (vf) is 0 ft/sec because the car comes to a complete stop. The initial velocity (vi) is 42 ft/sec, and the deceleration (a) is -7 ft/sec² (negative because the car is slowing down).
Plugging these values into the equation, we have:
0 ft/sec = 42 ft/sec + (-7 ft/sec²) * t
Rearranging the equation, we get:
7 ft/sec² * t = 42 ft/sec
Dividing both sides by 7 ft/sec², we find:
t = 6 sec
Now that we know the time it takes for the car to stop is 6 seconds, we can find the distance traveled using the formula:
Distance (d) = Initial velocity (vi) * Time (t) + 0.5 * Acceleration (a) * Time (t)²
Plugging in the values, we have:
d = 42 ft/sec * 6 sec + 0.5 * (-7 ft/sec²) * (6 sec)²
Simplifying the equation, we find:
d = 252 ft + (-126 ft)
d = 126 ft
Therefore, the car travels 126 feet before coming to a complete stop.
v = 42 - 7 t
when does v = 0?
t = 42/7
x = (1/2) a t^2
= (7/2)(42^2/49)
= 42^2/14
= 126 feet