Driving force over time to produce acceleration (graph)

In summary, a car of mass 1000 kg initially at rest moves along a straight road for 20 seconds before coming to rest again. The total distance traveled is 250 m. During the part of the motion represented by CD, the resultant force acting on the car is 4000 N. The momentum of the car when it reaches its maximum speed is 20 000 kg/ms, and the constant resultant accelerating force is 2000 N. Sketching a graph of the driving force varying with time to produce this constant acceleration would result in a horizontal line. However, this does not take into account air resistance, which would require a varying driving force to maintain a constant resultant force. When the car, traveling at its maximum speed
  • #1
moenste
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Homework Statement


Parts which I am not sure about are in bold.

(a) A car of mass 1000 kg is initially at rest. It moves along a straight road for 20 s and then comes to rest again. The speed-time graph for the movement is:
714ed3f62fa7.jpg


(i) What is the total distance travelled?
(ii) What resultant force acts on the car during the part of the motion represented by CD?
(iii) What is the momentum of the car when it has reached its maximum speed? Use this momentum value to find the constant resultant accelerating force.
(iv) During the part of the motion represented by OB on the graph, the constant resultant force found in (iii) is acting on the moving car although it is moving through air. Sketch a graph to show how the driving force would have to vary with time to produce this constant acceleration. Explain the shape of your graph.

(b) If, when traveling at this maximum speed, the 1000 kg car had struck and remained attached to a stationary vehicle of mass 1500 kg, with what speed would the interlocked vehicles have traveled immediately after collision?

Calculate the kinetic energy of the car just prior to this collision and the kinetic energy of the interlocked vehicles just afterwards. Common upon the values obtained.

Explain how certain design features in a modern car help to protect the driver of a car in such a collision.

2. The attempt at a solution
(a) (i) Total distance = 250 m.
(ii) F = m * (v / t) = 1000 * (20 / 5) = 4000 N
(iii) p = mv = 1000 * 20 = 20 000 kg / ms
F = p / t = 20 000 / 10 = 2000 N
(iv) Shouldn't the graph be a linear graph just a horizontal line? On the given graph we have velocity / time and that means that the acceleration is constant (velocity is increasing). Because of constant velocity, the driving force F = ma is constant as well (m and a don't change). So in that case I would draw the graph a horizontal line. Is this logic right?

(b) 20 * 1000 = v (1000 + 1500) -> v = 8 m / s
KE = 1/2 mv2 = 0.5 * 1000 * 202 = 200 000 J
KE = 0.5 * 2500 * 82 = 80 000 J

The car had 200 kJ before and the combined KE is 80 kJ. So we can conclude that there was a serious accident and 120 kJ have transformed into sound and head.

Seatbelts, airbags, bumpers? Or the whole car is made out of a better material so when there is an collision the car doesn't deform much and the KE doesn't transform into heat and sound?
 
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  • #2
For a iv), they want you to consider drag. But they have not told you how drag depends on speed, which is unusual. If you have not been taught anything about that, assume it is quadratic.

For the last part, you want a car to deform in a serious collision. That's how it absorbs the energy of the impact. But you don't want the passenger compartment to collapse, so cars have front and rear crumple zones. These are designed to provide a more-or-less constant resistance as they are crushed. That minimises the maximum forces on the passengers.
 
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  • #3
moenste said:

Homework Statement


Parts which I am not sure about are in bold.

(a) A car of mass 1000 kg is initially at rest. It moves along a straight road for 20 s and then comes to rest again. The speed-time graph for the movement is:
714ed3f62fa7.jpg


(i) What is the total distance travelled?
(ii) What resultant force acts on the car during the part of the motion represented by CD?
(iii) What is the momentum of the car when it has reached its maximum speed? Use this momentum value to find the constant resultant accelerating force.
(iv) During the part of the motion represented by OB on the graph, the constant resultant force found in (iii) is acting on the moving car although it is moving through air. Sketch a graph to show how the driving force would have to vary with time to produce this constant acceleration. Explain the shape of your graph.

(b) If, when traveling at this maximum speed, the 1000 kg car had struck and remained attached to a stationary vehicle of mass 1500 kg, with what speed would the interlocked vehicles have traveled immediately after collision?

Calculate the kinetic energy of the car just prior to this collision and the kinetic energy of the interlocked vehicles just afterwards. Common upon the values obtained.

Explain how certain design features in a modern car help to protect the driver of a car in such a collision.

2. The attempt at a solution
(a) (i) Total distance = 250 m.
(ii) F = m * (v / t) = 1000 * (20 / 5) = 4000 N
(iii) p = mv = 1000 * 20 = 20 000 kg / ms

Check your units on part (iii). Otherwise, so far so good.

F = p / t = 20 000 / 10 = 2000 N
(iv) Shouldn't the graph be a linear graph just a horizontal line? On the given graph we have velocity / time and that means that the acceleration is constant (velocity is increasing). Because of constant velocity, the driving force F = ma is constant as well (m and a don't change). So in that case I would draw the graph a horizontal line. Is this logic right?
I think it's asking you to account for air resistance. The "resultant force" is constant, and you have calculated that. What it's saying is that the "driving force" is not necessarily constant, and might vary to account for air resistance in order to keep the "resultant force" constant.

(b) 20 * 1000 = v (1000 + 1500) -> v = 8 m / s
KE = 1/2 mv2 = 0.5 * 1000 * 202 = 200 000 J
KE = 0.5 * 2500 * 82 = 80 000 J

The car had 200 kJ before and the combined KE is 80 kJ. So we can conclude that there was a serious accident and 120 kJ have transformed into sound and head.

Seatbelts, airbags, bumpers?
Those are all good answers. :smile:

Or the whole car is made out of a better material so when there is an collision the car doesn't deform much and the KE doesn't transform into heat and sound?

You might want to rethink that part though. The goal is to keep the driver and passengers from deforming (by keeping their acceleration to a minimum). It's not so bad if other parts of the car deform though.
 
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  • #4
haruspex said:
For a iv), they want you to consider drag. But they have not told you how drag depends on speed, which is unusual. If you have not been taught anything about that, assume it is quadratic.
I think it's asking you to account for air resistance. The "resultant force" is constant, and you have calculated that. What it's saying is that the "driving force" is not necessarily constant, and might vary to account for air resistance in order to keep the "resultant force" constant.
"Although it is moving through air" = so we need to consider air resistance.

But if we have Driving force F - Resistance R = Resultant force 2000 N and we have constant resistance (e.g. constant air resistance) should't the graph still be linear, a horizontal line of F? (if we consider that F, R and 2000 N are always constant). And in case the R is increasing over time then F also increases and the graph should be a line which increases like OB on the given graph. (over time we need more driving force to keep the resultant force the same).

I'm mostly not sure about the graph.

collinsmark said:
Check your units on part (iii). Otherwise, so far so good.
Yes, it should be kg m / s (kg m s-1).

For the last part, you want a car to deform in a serious collision. That's how it absorbs the energy of the impact. But you don't want the passenger compartment to collapse, so cars have front and rear crumple zones. These are designed to provide a more-or-less constant resistance as they are crushed. That minimises the maximum forces on the passengers.
You might want to rethink that part though. The goal is to keep the driver and passengers from deforming (by keeping their acceleration to a minimum). It's not so bad if other parts of the car deform though.
Hm, I meant that modern cars have good frames (e.g. like in a Smart car), so when there is a collusion these frames keep the car (the passenger part of the car) from completely deforming. But I understand it better after looking at the crumple zones which haruspex said. In regular vehicles the crumple zones are at the front and back of the car and the passenger part is protected with a frame.

So, in sum: in order to protect the passenger(s) in an accident, there are two types of protection. The first type is in the cabin: seatbelts and airbags which are deisgned to keep the passengers' acceleration to a minimum during a crash. And the second type are crumple zones with frames which allow the car to deform at the back and front but due to frames the passenger section remains relatively safe / untouched / unaffected.

---

P. S. [...] 120 kJ have transformed into sound and head -> heat, a typo :).
 
  • #5
moenste said:
But if we have Driving force F - Resistance R = Resultant force 2000 N and we have constant resistance (e.g. constant air resistance)
But that's exactly the point, air resistance is not constant. It depends on speed. No speed, no resistance. Typically, it is taken to rise quadratically with speed, but that's just an approximation and breaks down in some regimes.
 
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  • #6
haruspex said:
But that's exactly the point, air resistance is not constant. It depends on speed. No speed, no resistance. Typically, it is taken to rise quadratically with speed, but that's just an approximation and breaks down in some regimes.
So what's the correct graph in this situation?

One like this? (X = time, Y = driving force). This shape is due to air resistance which increases (as you say) quadratically with speed.
InvGBfilter.png
 
  • #7
moenste said:
So what's the correct graph in this situation?

One like this? (X = time, Y = driving force). This shape is due to air resistance which increases (as you say) quadratically with speed.
InvGBfilter.png
Very roughly something like that, but lose the units and axes legends, since you are not claiming a specific form of the function - just that it curves upwards.
 
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Related to Driving force over time to produce acceleration (graph)

1. What is the driving force over time to produce acceleration?

The driving force over time to produce acceleration refers to the amount of force applied to an object over a period of time in order to cause it to accelerate. This can be represented graphically as a line on a graph, with the force on the y-axis and time on the x-axis.

2. How is the driving force related to acceleration on a graph?

The driving force and acceleration are directly proportional on a graph. This means that as the driving force increases, the acceleration also increases, and vice versa.

3. What does a steep line on a driving force over time graph indicate?

A steep line on a driving force over time graph indicates a high rate of acceleration. This means that a large amount of force is being applied to the object over a short period of time, resulting in a significant change in its velocity.

4. How can I calculate the driving force from a graph?

To calculate the driving force from a graph, you can use the formula F = ma, where F is the driving force, m is the mass of the object, and a is the acceleration. Simply plug in the values from the graph and solve for F.

5. What factors can affect the driving force over time to produce acceleration?

The driving force over time to produce acceleration can be affected by a variety of factors, such as the mass of the object, the surface it is moving on, and any opposing forces (such as friction). The shape and angle of the force applied can also impact the acceleration of the object.

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