3.an object moves with constant acceleration 4.00 m/s2 and over a time interval reaches a final velocity of 12.0 m/s.. a. if its original velocity is 6.00 m/s, what is its displacement during the time interval? . b. what is the distance it travels during this interval? . c. if its original velocity is -6.00 m/s, what is its displacement during this interval? . d. what is the total distance it travels during the interval in part (

a) First of all, you need to find out the time. When a is constant, it can be calculated with this formula: a=(vf-vo)/t So you can find t like this: t=(vf-vo)/a=(12.00m/s-6.00m/s)/(4.00m/s^2)=1.50s There are several different methods to find the displacement knowing t. One of them uses the uniformly accelerated linear movement formula: d=1/2*a*t^2+vo*t=1/2*4.00m/s^2*(1.50s)^2+(6.00m/s)*1.50s=13.5m b) In this case, displacement and distance are the same thing. So the distance is also 13.5m. c) and d) This time displacement and distance are not the same thing. Take a look at the signs: Velocity starts being negative and later it becomes positive. This means the object is first going backwards while it is slowing down, and when it reaches v=0, it accelerates to the other side. Let’s find out how much time it needs to reach v=0 t=(vf-vo)/a=(0m/s-(-6.00m/s))/(4.00m/s^2)=1.50s d=1/2*4.00m/s^2*(1.50s)^2+(-6m/s)*1.50s=-4.5m So the distance during this period is -4.50m, but the displacement is always considered positive, so it is 4.50m. Let’s find out how much time it needs to reach v=12.0m/s t=(vf-vo)/a=(12.0m/s-(0m/s))/(4.00m/s^2)=3.00s Now the distance will be: d=1/2*4.00m/s^2*(3s)^2+(0m/s)*3.00s=18.0m And this time it is positive, so the displacement is also 18.0m. So the total displacement is: 4.50m+18m=22.5m And the total distance is: -4.50m+18m=13.5m There is another different method to calculate the distance, which doesn’t work for calculating the displacement (take a look at the signs and compare them). It consists on using just the initial velocity and the final velocity with the uniformly accelerated linear movement formula, without using 0 as an “intermediary”. t=(vf-vo)/a=(12.0m/s-(-6.00m/s))/(4.00m/s^2)=4.5s d=1/2*4.00m/s^2*(4.5s)^2+(-6m/s)*3.00s=22.5m You can see the result is equal to the distance we have calculated with the previous method, but it is not the same number we calculated for the displacement. Also keep in mind that each term of this last method is also different to each displacement we calculated before (before adding them).

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j