To analyze the motion of the car described in the problem, let's break it down into phases and then proceed to create the velocity-time graph, calculate accelerations/deceleration, and find the distance covered.
1. Phases of Motion:
The car has three main phases of motion:
- Phase 1: Acceleration from rest to 30 m/s in 15 seconds.
- Phase 2: Constant speed of 30 m/s for 30 seconds.
- Phase 3: Deceleration from 30 m/s to 0 m/s in 5 seconds.
2. Drawing the Velocity-Time Graph:
-
Phase 1:
- Start at (0, 0) and end at (15, 30).
- This will be a straight line sloping upward, indicating acceleration.
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Phase 2:
- From (15, 30) to (45, 30).
- This will be a horizontal line, indicating constant speed.
-
Phase 3:
- From (45, 30) to (50, 0).
- This will be a straight line sloping downward, indicating deceleration.
3. Calculating the Accelerations and Deceleration:
(a) Acceleration in Phase 1:
Using the formula: \[ a = \frac{\Delta v}{\Delta t} = \frac{30 , \text{m/s} - 0 , \text{m/s}}{15 , \text{s}} = \frac{30 , \text{m/s}}{15 , \text{s}} = 2 , \text{m/s}^2 \]
(b) Deceleration in Phase 3:
Using the formula: \[ a = \frac{\Delta v}{\Delta t} = \frac{0 , \text{m/s} - 30 , \text{m/s}}{5 , \text{s}} = \frac{-30 , \text{m/s}}{5 , \text{s}} = -6 , \text{m/s}^2 \]
4. Finding Total Distance Covered from the Velocity-Time Graph:
The area under the velocity-time graph represents distance. The graph consists of three shapes: a triangle, a rectangle, and another triangle.
(a) Area of Triangle (Phase 1):
\[ \text{Area}_{\text{Triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 15 , \text{s} \times 30 , \text{m/s = 225 , m} \]
(b) Area of Rectangle (Phase 2):
\[ \text{Area}_{\text{Rectangle}} = \text{length} \times \text{width} = 30 , \text{s} \times 30 , \text{m/s} = 900 , m \]
(c) Area of Triangle (Phase 3):
\[ \text{Area}_{\text{Triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 , \text{s} \times 30 , \text{m/s} = 75 , m \]
(d) Total Distance:
\[ \text{Total Distance} = \text{Distance}{\text{Phase 1}} + \text{Distance}{\text{Phase 2}} + \text{Distance}_{\text{Phase 3}} \] \[ \text{Total Distance} = 225 , m + 900 , m + 75 , m = 1200 , m \]
5. Verification using Formula Method:
To verify the distance covered during each phase:
(a) Distance during Phase 1:
Using the formula for distance during uniform acceleration: \[ d = ut + \frac{1}{2} a t^2 = 0 + \frac{1}{2} \cdot 2 , \text{m/s}^2 \cdot (15 , \text{s})^2 = 0.5 \cdot 2 \cdot 225 = 225 , m \]
(b) Distance during Phase 2:
For constant speed: \[ d = vt = 30 , \text{m/s} \cdot 30 , \text{s} = 900 , m \]
(c) Distance during Phase 3:
Using the formula for distance during uniform acceleration: \[ d = vt + \frac{1}{2} a t^2 = 30 , \text{m/s} \cdot 5 , \text{s} + \frac{1}{2} \cdot (-6) \cdot (5^2) \] \[ = 150 , m - 75 , m = 75 , m \]
Total calculated distance: \[ \text{Total Distance} = 225 , m + 900 , m + 75 , m = 1200 , m \]
Conclusion:
- Acceleration: \(2 , \text{m/s}^2\)
- Deceleration: \(-6 , \text{m/s}^2\)
- Total distance covered: \(1200 , m\) (verified by both methods).
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