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Math Problems Answer Key

1. Solution: Define the mass of the Earth to be x. Then, Mercury has a mass of 0.055x and Jupiter has a mass of 317.8x. Jupiter’s mass compared to Mercury is 317.8x/0.055x= 5778.18 times

2. Solution: The object is slowing down because the acceleration being negative means that the velocity is decreasing and decreasing in the positive direction means that it is slowing down. 

3. Solution: 9.58×16=153.28 seconds, which is 2:33.28 minutes, if Usain Bolt ran at the same speed as his 100m for 1600 meters. Then, it is a simple matter of calculation, where we do 3:43.13-2:33.28=1:09.85. 

4. Solution: It is simply 4h/6h, which is ⅔.

5. Solution: 4/30 hours= 0.133 hours= 8 minutes. The actual answer is 8 minutes and 20 seconds, but the times are approximated here.

6. Solution: This cannot be determined since we need to know the instantaneous rate of change of said object at 10.5s, while we only know the velocity at 0s and velocity of 11s. The acceleration does not always have to be constant, but if such was specified, the answer would be 1/11 m/s²

6b. Solution: First, we have to determine the values of c and k. As we are given that the initial velocity of the object is 3 m/s, we know that v(0)=3=d(0)²+(0)/22+c, which means that c=3. From here, we know that v(11)=4=d(11)²+11/22+3, so 121k+3+½=4, and k=1/242. Thus, v(t)=(t²/242)+(t/22)+3, and a(t)=(t/121)+(1/22). At 10.5 seconds,the acceleration is (10.5/121)+(1/22)=(21/242)+(11/242)=32/242=16/121 m/s². This is greater than the average acceleration on the interval from 0s to 11s, 1/11 m/s², as discussed in 6a, because the acceleration increases over time. 

6c. Solution: The average acceleration is equal to the average of the acceleration at those two points, which are 1/22 and 71/242, which is 41/242 m/s². Equating this with our formula gives that t=15 seconds. A trick for this could be to use the fact that because the acceleration is a linear equation, the average acceleration is equal to the instantaneous acceleration in between the two times, which is at 15s. 

7. Solution: At a time of x seconds, the position of the ball is 6x⁵+5x⁴+4x³+3x²+2x+1 seconds. In order to find the velocity of the ball at x seconds, we need to find the rate of change. This can be done through taking derivatives and direct substitution. The first derivative (velocity) of the position is 30x⁴+20x³+12x²+6x+2. Then, the acceleration, which is the second derivative, is 120x³+60x²+24x+6. 

8. Solution: When the truck is speeding up, it means that the absolute value of the velocity is increasing. Since the truck is moving backwards, that means that the velocity is already negative, so the velocity must be becoming “more negative”. Thus, the change in velocity is negative, so the acceleration is negative. 

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