A piano has been pushed to the top of the ramp at the back of a moving van. The workers think it is safe, but as they walk away, it begins to roll down the ramp.

If the back of the truck is 1.3 above the ground and the ramp is inclined at 30, how much time do the workers have to get to the piano before it reaches the bottom of the ramp?

1.26s

To find the time the workers have before the piano reaches the bottom of the ramp, we can use the kinematic equation for the motion along an inclined plane:

s = ut + (1/2)at^2

where:
s = distance traveled (height of the ramp)
u = initial velocity (0 as the piano starts from rest)
a = acceleration (due to gravity, which is approximately 9.8 m/s^2)
t = time taken

First, we need to find the distance traveled (s) along the ramp. We can use trigonometry to determine this:

s = hypotenuse * sin(angle)

Given:
height = 1.3 m
angle = 30 degrees

s = 1.3 * sin(30)
s = 1.3 * 0.5
s = 0.65 m

Now we can rearrange the kinematic equation to solve for time (t):

s = (1/2)at^2
0.65 = (1/2) * 9.8 * t^2
0.65 = 4.9 * t^2

Divide both sides by 4.9:

0.65 / 4.9 = t^2
0.132653 = t^2

Take the square root of both sides to find t:

t = √(0.132653)
t ≈ 0.364 s

Therefore, the workers have approximately 0.364 seconds to get to the piano before it reaches the bottom of the ramp.

To determine the time the workers have to get to the piano before it reaches the bottom of the ramp, we can use the principles of physics and the equations of motion.

First, let's consider the components of the piano's motion along the ramp. We can divide the piano's motion into two parts: motion along the ramp (in the direction of the inclined plane) and motion perpendicular to the ramp.

Along the ramp:
The piano moves under the influence of gravity, so its acceleration along the ramp, denoted as "a_ramp," can be calculated using the formula:

a_ramp = g * sin(theta)

Here, "g" represents the acceleration due to gravity (approximately 9.8 m/s^2), and "theta" represents the angle of inclination of the ramp (30 degrees in this case).

Perpendicular to the ramp:
The perpendicular motion of the piano does not contribute to its horizontal motion along the ramp. Since the workers will approach the piano from the side, this component of motion is independent of the piano's downward motion along the ramp. Hence, we can ignore this component for now.

To find the time the workers have, we need to calculate the time it takes for the piano to reach the bottom of the ramp. We can use the formula of motion:

s = ut + (1/2) * a * t^2

Where:
- "s" is the distance traveled along the ramp (equal to the ramp length).
- "u" is the initial velocity along the ramp (assumed to be zero since the piano starts from rest).
- "a" is the acceleration along the ramp (which we calculated earlier as g * sin(theta)).
- "t" is the time taken to travel the distance "s."

In this case, the distance traveled along the ramp is equal to the height of the back of the truck, which is given as 1.3 meters.

By rearranging the equation, we can solve for "t":

t = sqrt(2s / a)

Substituting the known values:

t = sqrt(2 * 1.3 / (g * sin(theta)))

t = sqrt(2 * 1.3 / (9.8 * sin(30)))

t ≈ sqrt(0.2663 / 4.9)

t ≈ sqrt(0.054414)

t ≈ 0.2334 seconds (approximately)

Therefore, the workers have approximately 0.2334 seconds to get to the piano before it reaches the bottom of the ramp.