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a)$2.00 g's$
b)$9.52 \times 10^3 \textrm{N}$

Giancoli 6th Edition, Chapter 4, Problem 11


Chapter 4, Problem 11 is solved.

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Transcript for this Giancoli solution

This car we'll assume has an initial velocity of zero meters per second and it covers four hundred two meters in six point four seconds. So our job is to figure out the acceleration in 'g's' which we'll do using this kinematics formula: 'd' equals 'vit' plus one half 'at' squared. The initial velocity is zero which makes this first term go away, and we'll solve this for 'a' by multiplying both sides by two and dividing by 't' squared. Times two and divide by 't' squared on the left as well. Flip the sides around giving us that 'a' is two 'd' over 't' squared and equals two times four hundred two meters divided by six point four seconds squared. This gives nineteen point six three meters per second squared, which is not our final answer because we're told to express it in units of 'g's'. So we have nineteen point six three meters per second squared times one 'g' for every nine point eight meters per second squared. And these meters per second squareds will cancel so we're really going nineteen point six three divided by nine point eight which gives two point zero zero g's for the 6th Edition. That's the acceleration. The force is, well there is only one force horizontally so that one force is going to be the net force, and for that reason we can call it 'ma'. It's going to be four hundred eighty five kilograms times nineteen point six three meters per second squared. And that gives nine point five two times ten to the three Newtons is the force that the road applies on the tires of the car.