A toy slides down an inclined plane which makes an angle of 60 degrees with the horizontal. It starts from rest and achieves a final velocity of 8 m/s in 5 seconds at the bottom of the plane.
a. What is the acceleration down the plane?
b. What is the distance along the plane it travels before reaching the bottom of the plane?
a. V = a*t = 8 m/s
a*5 = 8
a = 1.6 m/s^2.
b. d = 0.5a*t^2
To find the answers to these questions, we can use the equations of motion and trigonometry.
a. To find the acceleration down the plane, we can use the equation of motion:
v = u + at,
where:
v = final velocity = 8 m/s,
u = initial velocity = 0 m/s (as it starts from rest),
a = acceleration, and
t = time taken = 5 seconds.
Rearranging the equation, we get:
a = (v - u) / t.
Substituting the given values, we have:
a = (8 m/s - 0 m/s) / 5 s,
a = 8 m/s / 5 s,
a = 1.6 m/s².
Therefore, the acceleration down the plane is 1.6 m/s².
b. To find the distance along the plane it travels before reaching the bottom, we can use trigonometry. We can use the perpendicular component of gravity (mg sin θ), where m is the mass of the toy and g is the acceleration due to gravity (approximately 9.8 m/s²).
The equation we can use is:
Distance (d) = (1/2) * a * t²,
where:
a = acceleration = 1.6 m/s² (as found in Part a),
t = time taken = 5 seconds.
Substituting the values, we have:
d = (1/2) * 1.6 m/s² * (5 s)²,
d = (1/2) * 1.6 m/s² * 25 s²,
d = 20 m.
Therefore, the distance along the plane the toy travels before reaching the bottom is 20 meters.