1) A boy sledding down a hill accelerates at 1.40 m/s, If he started at rest, in what distance would he reach a speed of 7.00 m/s?
2) An Elevator is moving upward at 1.20 m/s when it experiences an accelerstion of 0.31 m/s downward, over a distance of 0.75m. What will its final velocity be?
17.5
To answer these questions, we can use the equations of motion. There are four equations of motion that relate the final velocity (v), initial velocity (u), acceleration (a), distance (s), and time (t). The equations are as follows:
1) v^2 = u^2 + 2as
2) v = u + at
3) s = ut + (1/2)at^2
4) s = vt - (1/2)at^2
Let's solve the first question:
1) We are given:
Initial velocity (u) = 0 m/s (since the boy starts at rest)
Acceleration (a) = 1.40 m/s^2
Final velocity (v) = 7.00 m/s
We need to find the distance (s) covered by the boy. We can rearrange equation 1 and solve for s:
v^2 = u^2 + 2as
(7.00 m/s)^2 = (0 m/s)^2 + 2 * 1.40 m/s^2 * s
49.00 m^2/s^2 = 0 m^2/s^2 + 2.80 m/s^2 * s
49.00 m^2/s^2 = 2.80 m/s^2 * s
Solving for s, we have:
s = (49.00 m^2/s^2) / (2.80 m/s^2)
s ≈ 17.50 m
Therefore, the boy will reach a speed of 7.00 m/s in approximately 17.50 meters.
Now let's move on to the second question:
2) We are given:
Initial velocity (u) = 1.20 m/s (upward)
Acceleration (a) = -0.31 m/s^2 (downward)
Distance (s) = 0.75 m
We need to find the final velocity (v). Again, we can rearrange equation 1 and solve for v:
v^2 = u^2 + 2as
v^2 = (1.20 m/s)^2 + 2 * (-0.31 m/s^2) * 0.75 m
v^2 ≈ 1.44 m^2/s^2 - 0.4656 m^2/s^2
v^2 ≈ 0.9744 m^2/s^2
Taking the square root of both sides, we have:
v ≈ √(0.9744 m^2/s^2)
v ≈ 0.987 m/s (rounded to three significant figures)
Therefore, the final velocity of the elevator will be approximately 0.987 m/s (downward).
1) To find the distance the boy would reach a speed of 7.00 m/s, we can use the equation for acceleration:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.
In this case, the initial velocity (u) is 0 m/s, the acceleration (a) is 1.40 m/s^2, and the final velocity (v) is 7.00 m/s. Plugging these values into the equation:
7.00^2 = 0^2 + 2 * 1.40 * s
Simplifying:
49.00 = 2.80s
Dividing both sides by 2.80:
s = 17.50 m
Therefore, the boy would reach a speed of 7.00 m/s in a distance of 17.50 meters.
2) To find the final velocity of the elevator, we can use the equation:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
In this case, the initial velocity (u) is 1.20 m/s, the acceleration (a) is -0.31 m/s^2 (negative because it's downward), and the distance (s) is 0.75 m. We need to solve for time.
We can use the equation:
s = ut + (1/2)at^2
Since the initial velocity is 1.20 m/s, the equation becomes:
0.75 = 1.20t + (1/2)(-0.31)t^2
Simplifying:
0.75 = 1.20t - 0.155t^2
Rearranging:
0.155t^2 - 1.20t + 0.75 = 0
Now we can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
where a = 0.155, b = -1.20, and c = 0.75. Plugging in these values:
t = (-(-1.20) ± √((-1.20)^2 - 4(0.155)(0.75))) / (2(0.155))
Simplifying:
t = (1.20 ± √(1.44 - 0.465)) / (0.31)
t = (1.20 ± √0.975) / (0.31)
Calculating √0.975:
t = (1.20 ± 0.987) / (0.31)
t = (1.20 + 0.987) / (0.31) or t = (1.20 - 0.987) / (0.31)
Simplifying:
t = 4.13 s or t = 0.76 s
Since the elevator is moving upward, we can disregard the negative solution. Therefore, the time taken is 0.76 seconds.
Now we can plug this value back into the equation for final velocity:
v = u + at
v = 1.20 + (-0.31)(0.76)
v = 1.20 - 0.2356
v = 0.9644 m/s
Therefore, the final velocity of the elevator would be approximately 0.96 m/s.