A vehicle starts from rest and in 30s reaches a speed of 20m/s. Thereafter the speed remains steady for 15s and decreases steadily until it stops 5s.
i. Use the data provided to draw a velocity-time graph for the motion
ii. Calculate the acceleration during the first 30s
iii. What is the total distance covered by the vehicle
(i) the graph consists of three straight-line segments, where the slope of each section is the acceleration.
(ii) a = 20/30 m/s^2
the graph will show the answer to (iii)
i. To draw the velocity-time graph, we can break down the motion into three parts: the initial acceleration, the constant speed, and the deceleration.
1. Initial acceleration: The vehicle starts from rest and reaches a speed of 20 m/s in 30 seconds. This means the initial acceleration can be calculated using the equation:
Acceleration = (Final Velocity - Initial Velocity) / Time
Acceleration = (20 m/s - 0 m/s) / 30 s = 0.67 m/s^2
2. Constant speed: The speed remains steady at 20 m/s for 15 seconds. This segment of the graph will be a flat horizontal line at 20 m/s.
3. Deceleration: The speed decreases steadily until it stops in 5 seconds. Since the vehicle comes to a stop, we know the final velocity is 0 m/s. Using the equation for acceleration, we can calculate the deceleration:
Acceleration = (Final Velocity - Initial Velocity) / Time
Acceleration = (0 m/s - 20 m/s) / 5 s = -4 m/s^2 (negative sign indicates deceleration)
Combining these three parts, the velocity-time graph will look like:
^
Speed | -
(m/s) | -
-
-
-
-
-
----------------------------------> Time (s)
ii. The acceleration during the first 30 seconds is 0.67 m/s^2.
iii. To find the total distance covered by the vehicle, we need to calculate the areas under each segment of the velocity-time graph.
1. For the initial acceleration segment, the area under the graph represents the distance covered. To find this area, we can use the equation:
Distance = (Initial Velocity * Time) + (0.5 * Acceleration * Time^2)
Distance = (0 * 30) + (0.5 * 0.67 * 30^2) = 301.5 m
2. For the constant speed segment, the area under the graph is a rectangle, and the distance covered is given by:
Distance = Speed * Time
Distance = 20 m/s * 15 s = 300 m
3. For the deceleration segment, the area under the graph represents the distance covered. Similar to the initial acceleration segment, we can use the equation:
Distance = (Initial Velocity * Time) + (0.5 * Acceleration * Time^2)
Distance = (20 * 5) + (0.5 * -4 * 5^2) = 95 m
Therefore, the total distance covered by the vehicle is:
301.5 m + 300 m + 95 m = 696.5 m
To draw the velocity-time graph and calculate the acceleration, we need to analyze the given information step by step.
i. Drawing the velocity-time graph:
- The vehicle starts from rest, so the initial velocity is 0 m/s.
- In the first 30s, the vehicle reaches a speed of 20m/s. This means the slope of the graph will be a constant positive value during this period.
- After 30s, the speed remains steady at 20m/s. Therefore, the graph will be a straight horizontal line indicating a constant velocity.
- Finally, the vehicle decreases steadily until it stops in 5s. This means the slope of the graph will be a constant negative value during this period.
ii. Calculating the acceleration during the first 30s:
Acceleration is the rate at which the velocity changes over time. In this case, we can calculate the acceleration during the first 30s using the formula:
acceleration = (final velocity - initial velocity) / time
Given:
Initial velocity (u) = 0 m/s
Final velocity (v) = 20 m/s
Time (t) = 30s
Substituting the values into the formula:
acceleration = (20 m/s - 0 m/s) / 30s
acceleration = 20 m/s / 30s
acceleration ≈ 0.67 m/s²
The acceleration during the first 30s is approximately 0.67 m/s².
iii. Calculating the total distance covered by the vehicle:
To calculate the total distance covered, we need to break the motion into different intervals.
Interval 1: The vehicle accelerates from rest to 20 m/s in 30s.
In this interval, we can calculate the distance covered using the formula:
distance = (initial velocity * time) + (0.5 * acceleration * time²)
Given:
Initial velocity (u) = 0 m/s
Final velocity (v) = 20 m/s
Time (t) = 30s
Acceleration (a) ≈ 0.67 m/s²
Substituting the values into the formula:
distance = (0 m/s * 30s) + (0.5 * 0.67 m/s² * (30s)²)
distance = 0m + 0.5 * 0.67 m/s² * 900s²
distance ≈ 301.5m
Interval 2: The vehicle maintains a constant speed of 20 m/s for 15s.
In this interval, the distance covered can be calculated using the formula:
distance = velocity * time
Given:
Velocity (v) = 20 m/s
Time (t) = 15s
Substituting the values into the formula:
distance = 20 m/s * 15s
distance = 300m
Interval 3: The vehicle decreases in speed until it stops in 5s.
Since the speed decreases uniformly, the distance can be calculated using the formula:
distance = (initial velocity * time) - (0.5 * acceleration * time²)
Given:
Initial velocity (u) = 20 m/s
Final velocity (v) = 0 m/s
Time (t) = 5s
Acceleration (a) ≈ 0.67 m/s² (negative sign as it's decelerating)
Substituting the values into the formula:
distance = (20 m/s * 5s) - (0.5 * 0.67 m/s² * (5s)²)
distance = 100m - 0.5 * 0.67 m/s² * 25s²
distance = 100m - 0.5 * 0.67 m/s² * 625s²
distance ≈ 100m - 209.375m
distance ≈ -109.375m (negative because it's in the opposite direction)
To calculate the total distance covered by the vehicle, we sum up the distances from each interval:
Total distance = distance in interval 1 + distance in interval 2 + distance in interval 3
Total distance = 301.5m + 300m - 109.375m
Total distance ≈ 492.125m
Therefore, the total distance covered by the vehicle is approximately 492.125 meters.