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Hi guys, im really stuck on this question:

A car of mass 1050kg moves along a straigth horizontal road with its engine working at a constant rate of 25kW. Its speed at a point A on the road is 12ms^-1. Assuming that there is no resisitance to motion, calculate the time taken for the car ti travel from A until it reaches a speed of 20ms^-1

Assume now there is a constant resistance to motion and that the car's engine continues to work at 25kW. It takes 10.7s for the car's speed to increase from 12ms^-1 to 20ms^-1. During this time the car travels 179m. Calculate the work done against the resistance and hence find the magnitude of the resistance

LAter the car moves up a straight hill, inclined at 2 degrees to the horizontal. The engine works at 25kW as before, and there is a constant resistance of the same magnitude as before. The car travels a distance of 393m while its speed increases from 12ms^-1 to 20ms^-1. Calculate the time taken by the car to travel this distance.

book says the answers are: 5.38s, 133kJ, 744N, 22.7s

thanx in advance, i know im missing something simple, please enlighten me.

A car of mass 1050kg moves along a straigth horizontal road with its engine working at a constant rate of 25kW. Its speed at a point A on the road is 12ms^-1. Assuming that there is no resisitance to motion, calculate the time taken for the car ti travel from A until it reaches a speed of 20ms^-1

Assume now there is a constant resistance to motion and that the car's engine continues to work at 25kW. It takes 10.7s for the car's speed to increase from 12ms^-1 to 20ms^-1. During this time the car travels 179m. Calculate the work done against the resistance and hence find the magnitude of the resistance

LAter the car moves up a straight hill, inclined at 2 degrees to the horizontal. The engine works at 25kW as before, and there is a constant resistance of the same magnitude as before. The car travels a distance of 393m while its speed increases from 12ms^-1 to 20ms^-1. Calculate the time taken by the car to travel this distance.

book says the answers are: 5.38s, 133kJ, 744N, 22.7s

thanx in advance, i know im missing something simple, please enlighten me.

actually dont worry about the first part, i just worked it out

Ahoy Ahoy

Second part of the question, let us first find the work done against no resistance = P*t = 25000*5.38 = 134500 J

Then, using new value of 10.7 as time, total Work done = 25000*10.7 = 267500 J

Subtract the first from the second the find the work done against resistance at 133Kj

From this, one can work out the magnitude of force against resistance;

133000 = WD = F*d = F*179 => F=133000/179 = 743 Newtons

For the third part, consider the SUVAT equation to work out acceleration

V²=u²+2as

=> 400 = 144 + 786a

=> a= 0.326

=>Resultant force = ma = 342N

Total force required = 342 - (-743) - (-9.8*1050*sin2) = 1444 N

=> 1444*393/t = 25000

=> 1444*393/25000 = t

=> t = 22.7 seconds

Hope this helps, don't know about the 743N instead of 744N though ;/

Second part of the question, let us first find the work done against no resistance = P*t = 25000*5.38 = 134500 J

Then, using new value of 10.7 as time, total Work done = 25000*10.7 = 267500 J

Subtract the first from the second the find the work done against resistance at 133Kj

From this, one can work out the magnitude of force against resistance;

133000 = WD = F*d = F*179 => F=133000/179 = 743 Newtons

For the third part, consider the SUVAT equation to work out acceleration

V²=u²+2as

=> 400 = 144 + 786a

=> a= 0.326

=>Resultant force = ma = 342N

Total force required = 342 - (-743) - (-9.8*1050*sin2) = 1444 N

=> 1444*393/t = 25000

=> 1444*393/25000 = t

=> t = 22.7 seconds

Hope this helps, don't know about the 743N instead of 744N though ;/

Kaiser mole, u r a saviour, thnx sooo much

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- Edexcel IAL Maths & Further Mathematics
- nuclear physics inverse square law
- 1st year of uni- iPad dilemma
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- Deciding Edexcel IAL FM Cash-in Unit Combinations
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