1.2 Motion (3)
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1.
An object starts from rest and accelerates at a constant rate of 2.5 m/s2. The speed-time graph for this motion is shown below.
Determine, from the graph, the following:
- The acceleration of the object.
- The distance travelled by the object in the first 5 seconds.
- The time taken for the object to reach a speed of 10 m/s.
(a) Constant Acceleration:
The acceleration is constant because the slope of the speed-time graph is constant. From the graph, the slope is 2.5 m/s2. Therefore, the acceleration is 2.5 m/s2.
(b) Distance Travelled:
The distance travelled is represented by the area under the speed-time graph. In this case, the area is a triangle. The area of a triangle is (1/2) * base * height. The base is 5 seconds and the height is the speed at 5 seconds, which is 12.5 m/s. Therefore, the distance travelled is (1/2) * 5 s * 12.5 m/s = 31.25 m. So, the distance travelled in the first 5 seconds is 31.25 m.
(c) Time to Reach 10 m/s:
The object reaches a speed of 10 m/s when the speed-time graph intersects the line speed = 10 m/s. From the graph, this occurs at approximately 2 seconds. Therefore, the time taken for the object to reach a speed of 10 m/s is 2 seconds.
2.
A student measures the weight of an object on Earth to be 49 N. The student then travels to the Moon, where the acceleration due to gravity is approximately 1/6th of that on Earth. Calculate the mass of the object. Show your working.
We know that weight (W) = mass (m) * acceleration due to gravity (g). We are given:
- W = 49 N
- g (on the Moon) = (1/6) * g (on Earth)
- g (on Earth) ≈ 9.8 m/s2
Therefore, g (on the Moon) ≈ (1/6) * 9.8 m/s2 ≈ 1.63 m/s2
We can rearrange the formula to solve for mass (m):
Substituting the given values:
m = 49 N / 1.63 m/s2 ≈ 30.0 kg
Therefore, the mass of the object is approximately 30.0 kg.
3.
A rocket is accelerating upwards. The speed-time graph for the rocket's motion is shown below.
Determine, from the graph, the following:
- The acceleration of the rocket between t = 0s and t = 3s.
- The distance the rocket travels between t = 0s and t = 3s.
- The time taken for the rocket to reach a speed of 200 m/s.
(a) Acceleration between 0s and 3s:
The acceleration is the gradient of the speed-time graph. The change in speed is 200 m/s - 0 m/s = 200 m/s and the change in time is 3s - 0s = 3s. Therefore, the acceleration is (200 m/s) / (3s) = 66.67 m/s2 (approximately).
(b) Distance Travelled between 0s and 3s:
The distance travelled is the area under the speed-time graph. This area is a trapezium. The parallel sides are 0 m/s and 200 m/s, and the distance between them is 3s. The area of a trapezium is (1/2) * (sum of parallel sides) * height. Therefore, the area is (1/2) * (0 m/s + 200 m/s) * 3s = (1/2) * 200 m/s * 3s = 300 m.
(c) Time to Reach 200 m/s:
The rocket reaches a speed of 200 m/s when the speed-time graph intersects the line speed = 200 m/s. From the graph, this occurs at approximately 1.5 seconds. Therefore, the time taken for the rocket to reach a speed of 200 m/s is 1.5 seconds.