1. (a) A frictionless car of mass 1.5 tonnes travels uphill along a slope at 20º to the horizontal for 0.5 km at 20 km per hour.
(i) How much work does the engine do on the car?
(ii) What power does the engine supply?
(b) The car the rolls back down the hill with the engine turned off. How fast is it travelling when it has rolled back to its starting position?
(c) The car continues to roll a further 0.5 km down the slope until it hits a solid wall, whereupon it is brought to rest in a distance of 20 cm as the front of the car crumples.
(i) What impulse is applied to the wall?
(ii) Assuming uniform deceleration, what is the average braking force applied to the car by the wall?
(d) If the driver had noticed the wall when the car had rolled back as far as the starting point and had applied the brakes at that point, what average braking force and power would have been required to stop the car just as it reached the wall?
2. A car travels northeastwards for 2.5 km, accelerating with constant acceleration from stationary to 80 km per hour over the first 0.4 km and then travelling at constant velocity and finally breaking with constant deceleration to 5 km per hour over the final 0.2 km, at which point it turns 65º to the south. It then travels for another 3.7 km, initially accelerating to 65 km per hour for 0.28 km and then travelling at constant velocity before finally coming to rest by braking with constant deceleration for 10 s.
(a) Represent this journey accurately on distance-time, speed-time and acceleration-time graphs.
(b) Calculate the total distance and final displacement, the time taken for the journey, the average speed and the average velocity.
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To solve this problem, we need to apply the relevant formulas and equations. Let's break down each part of the question step-by-step:
1. (a)
(i) The work done by the engine can be calculated using the formula: work = force x distance x cos(theta). In this case, the force is equal to the weight of the car, which is the product of mass and acceleration due to gravity (m*g). Since the car is traveling uphill, the angle theta is given as 20º. The distance is 0.5 km, which needs to be converted to meters. The speed of the car is given as 20 kph, which needs to be converted to m/s. Therefore, you can calculate the work done by the engine as follows:
mass = 1.5 tonnes = 1500 kg
g = 9.8 m/s^2
distance = 0.5 km = 500 m
speed = 20 kph = 5.56 m/s
Now, calculate the work done:
work = force x distance x cos(theta)
force = mass x acceleration
acceleration = (speed^2) / distance
Substitute the values:
force = (mass x speed^2) / distance = (1500 x 5.56^2) / 500
work = force x distance x cos(theta)
(ii) The power supplied by the engine can be calculated using the formula: power = work / time. In this case, we need to calculate the time taken to travel the given distance at the given speed. To do this, you'll need to use the formula: time = distance / speed.
Substitute the values into the formula:
time = distance / speed = 500 / 5.56
Finally, substitute the calculated work and time into the power formula:
power = work / time
(b) When the car rolls back down the hill with the engine turned off, the main idea is that the car has the same potential energy at the starting position as at the ending position (ignoring any energy losses). This potential energy is converted into kinetic energy. Therefore, you can equate the potential energy at the starting position to the kinetic energy at the ending position:
m*g*h = (1/2) * m * v^2
Since the car is at the same height as before, h = 0, so the equation simplifies to:
0 = (1/2) * m * v^2
Solve for v:
v = 0
Therefore, the car is not moving when it has rolled back to its starting position.
(c) (i) The impulse applied to the wall can be calculated using the formula: impulse = change in momentum. Since the car is brought to rest, the change in momentum is equal to the initial momentum:
impulse = initial momentum = mass x velocity
Substitute the given values:
mass = 1.5 tonnes = 1500 kg
velocity = 20 m/s
(ii) To calculate the average braking force applied to the car by the wall, you can use the formula: force = mass x acceleration. In this case, the acceleration is given as uniform deceleration, so it is equal to -ve^2 / (2d), where e is the speed at impact and d is the distance traveled until the car stops.
Substitute the given values:
mass = 1500 kg
e = 20 m/s (since the car is brought to rest)
d = 20 cm = 0.2 m
Now, calculate the acceleration:
acceleration = -v^2 / (2d) = -(20^2) / (2 * 0.2)
(d) If the driver had noticed the wall when the car had rolled back to the starting point and had applied the brakes at that point, you can calculate the average braking force and power required to stop the car just as it reached the wall using the same formulas as in part (c)(ii):
force = mass x acceleration
power = force x speed
Use the given values:
mass = 1500 kg
speed = 20 m/s
acceleration is calculated using the same formula as in part (c)(ii)
2. (a) To accurately represent the journey on distance-time, speed-time, and acceleration-time graphs, you will need to analyze each segment of the journey. Plot the distance or displacement on the y-axis and time on the x-axis. Similarly, plot speed on the y-axis and time on the x-axis. For the acceleration-time graph, plot acceleration on the y-axis and time on the x-axis.
Break down the journey into segments and analyze each one individually:
Segment 1: Accelerating from stationary to 80 kph over the first 0.4 km.
Segment 2: Traveling at constant velocity.
Segment 3: Decelerating from 80 kph to 5 kph over the final 0.2 km.
Segment 4: Turning 65º to the south.
Segment 5: Traveling for another 3.7 km.
Segment 6: Accelerating to 65 kph for 0.28 km.
Segment 7: Traveling at constant velocity.
Segment 8: Coming to rest by braking with constant deceleration for 10 s.
Analyze each segment, determine the relevant equations, and plot the graphs accordingly.
(b) To calculate the total distance, final displacement, time taken for the journey, average speed, and average velocity, analyze each segment of the journey and use the relevant formulas:
Total distance = Sum of individual segment distances
Final displacement = Displacement from the starting point to the ending point
Time taken for the journey = Sum of individual segment times
Average speed = Total distance / Time taken for the journey
Average velocity = Final displacement / Time taken for the journey
Analyze each segment, calculate the relevant values, and plug them into the formulas to get your final results.