A race car can be slowed with a constant acceleration of -14 m/s2.
(a) If the car is going 44 m/s, how many meters will it travel before it stops?
(b) How many meters will it take to stop a car going twice as fast?
44/14
what would part b be?
(a) To find the distance the car will travel before stopping, we can use the equation:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s, as the car stops)
u = initial velocity (44 m/s)
a = acceleration (-14 m/s^2, negative as it's slowing down)
s = distance
Rearranging the equation to solve for s:
s = (v^2 - u^2) / (2a)
Substituting the values:
s = (0^2 - 44^2) / (2 * -14) = -1936 / -28 = 69.14 meters
Therefore, the race car will travel approximately 69.14 meters before it stops.
(b) If the car is going twice as fast, its initial velocity will be 2 * 44 = 88 m/s.
Using the same equation:
s = (0^2 - 88^2) / (2 * -14) = -7744 / -28 = 276.57 meters
Therefore, it will take approximately 276.57 meters to stop a car going twice as fast.
To solve this problem, we need to use the equations of motion.
(a) If the car is going 44 m/s, we need to find the distance it will travel before it stops. First, we need to find the time it takes for the car to stop.
We can use the equation of motion:
v = u + at,
where:
v = final velocity (0 m/s as the car stops)
u = initial velocity (44 m/s)
a = acceleration (-14 m/s², negative because it is slowing down)
t = time
Rearranging the equation to solve for time, we get:
t = (v - u) / a.
Substituting the given values, we have:
t = (0 - 44) / (-14)
t = 44 / 14
t = 3.14 seconds (approximately)
Now that we have the time, we can find the distance traveled using another equation of motion:
s = ut + (1/2)at²,
where:
s = distance traveled
t = time (3.14 seconds)
u = initial velocity (44 m/s)
a = acceleration (-14 m/s²)
Substituting the values, we get:
s = (44)(3.14) + (0.5)(-14)(3.14)²
s = 137.36 - 65.72
s = 71.64 meters (approximately)
Therefore, the car will travel approximately 71.64 meters before it stops.
(b) If the car is going twice as fast, its initial velocity would be 2 * 44 = 88 m/s.
Using the same equation, v = u + at, we can find the time it takes for the car to stop:
t = (0 - 88) / (-14)
t = 88 / 14
t = 6.28 seconds (approximately)
Again, using the equation s = ut + (1/2)at², we can find the distance traveled:
s = (88)(6.28) + (0.5)(-14)(6.28)²
s = 552.64 - 132.88
s = 419.76 meters (approximately)
Therefore, it will take approximately 419.76 meters to stop a car going twice as fast.