1) A car traveling at 22m/s decelerates at a constant 1.5m/s^2. Calculate the time it takes to stop.
I first calculated the distance which is 160 m. Then I used the formula d=vit+1/2at^2. I got a quadratic -0.75t^2+22t-160. I got 13 or 16 as my time, but the answer says 15. Would it be correct to take the average of those times?
2) A stone is dropped from the roof of a high building. A second stone is dropped 1s later. How far apart are the stones when the second one has reached a speed of 23.0m/s?
I set this problem up by writing the given for stone 1 and stone 2.
Some equations I got were d(s1)= 4.905t^2 and ts2=ts1+1
I don't know how to do the problem.
1. No, not the average.
vf=vi+at
0=22-1.5t
t=15seconds.
Your method rounded distance to 160m. If you had not rounded, you would have gotten 15 seconds....
2.find the time of the first to get to that speed.
vf=gt solve for t, and the distance fell.
Now, subtract one from that t, and find out how far the second fell.
Subtract the distances.
1) To calculate the time it takes for the car to stop, we can use the equation of motion:
v = u + at,
where:
- v is the final velocity (in this case, 0 m/s),
- u is the initial velocity (22 m/s),
- a is the deceleration (1.5 m/s^2), and
- t is the time taken.
Initially, the car is moving with a positive velocity (forward direction), so to account for the deceleration, we need a negative acceleration.
Rearranging the equation, we get:
0 = 22 + (-1.5)t.
Simplifying further:
-1.5t = -22,
t = -22 / -1.5,
t = 14.67 seconds (approximately).
So, based on the calculation, the car takes approximately 14.67 seconds to come to a stop.
Now, let's address your second question.
2) To determine the distance between the two stones when the second one has reached a speed of 23.0 m/s, we need to find the time it takes for the second stone to reach that speed and then calculate the distance traveled by both stones during that time.
Let's label the time when the first stone is dropped as t1 (t1 = 0) and the time when the second stone is dropped as t2 (t2 = 1).
For the first stone (stone 1), the equation of motion is given by:
d(s1) = 0.5 * g * t^2,
where:
- d(s1) represents the distance traveled by the first stone (or its height from the roof),
- g is the acceleration due to gravity (approximately 9.8 m/s^2),
- t is the time elapsed for the first stone.
For the second stone (stone 2), we can also use the same equation:
d(s2) = 0.5 * g * (t - 1)^2,
where:
- d(s2) represents the distance traveled by the second stone (or its height from the roof),
- t - 1 is the time elapsed for the second stone (since it was dropped 1 second later).
Now, we can set up another equation using the given condition:
23 = g * (t - 1).
From the equation for stone 2, we can express t as:
t = sqrt(2 * d(s2) / g) + 1.
Substituting this expression into the equation for stone 2, we can solve for d(s2):
23 = g * (sqrt(2 * d(s2) / g) + 1 - 1),
23 = sqrt(2 * d(s2) * g) / sqrt(g),
23 * sqrt(g) = sqrt(2 * d(s2) * g),
23^2 * g = 2 * d(s2) * g,
23^2 = 2 * d(s2),
d(s2) = (23^2) / 2,
d(s2) = 264.5 meters (approximately).
So, when the second stone reaches a speed of 23.0 m/s, the distance between the two stones is approximately 264.5 meters.
If you have any further questions, feel free to ask!