A passenger on an evacuation slide starts from rest at the top of a 45° slope. If thetrip to the bottom takes 3.6 s, how long is the slope? Assume that frictional forcesmay be neglected.A. 90 mB. 45 mC. 1130 mD. Cannot be calculated without knowing the passenger’s mass.[1 mark]
Question
A passenger on an evacuation slide starts from rest at the top of a 45° slope. If thetrip to the bottom takes 3.6 s, how long is the slope? Assume that frictional forcesmay be neglected.A. 90 mB. 45 mC. 1130 mD. Cannot be calculated without knowing the passenger’s mass.[1 mark]
Solution 1
To solve this problem, we can use the equations of motion. Since the passenger starts from rest, the initial velocity (u) is 0. The angle of the slope is 45°, so the acceleration (a) due to gravity is g*sin(45°), where g is the acceleration due to gravity (9.8 m/s²). The time (t) is given as 3.6 seconds. We need to find the distance (s) which is the length of the slope.
We can use the equation of motion: s = ut + 0.5at²
Substituting the given values:
s = 03.6 + 0.59.8*sin(45°)*3.6²
s = 0 + 0.59.80.707*12.96
s = 45.36 m
So, the length of the slope is approximately 45 m. The correct answer is B. 45 m.
Solution 2
To solve this problem, we can use the equations of motion. Since the passenger starts from rest, the initial velocity (u) is 0. The angle of the slope is 45°, so the acceleration (a) due to gravity will be g*sin(45°), where g is the acceleration due to gravity (9.8 m/s²). The time (t) is given as 3.6 seconds.
We can use the equation of motion: s = ut + 0.5at²
Substituting the given values:
s = 03.6 + 0.59.8*sin(45°)*3.6²
Solving this equation will give us the length of the slope.
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