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It’s due by tonight, please show work.

It’s due by tonight, please show work.-example-1

1 Answer

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First, you’re gonna need to know the basic formulas for velocity and acceleration. You’re also gonna need to know the kinetic formulas it seems like.
v = change in position/change in time
a = change in velocity/change in time
Vf = Vi + at
Vf^2 = Vi^2 + 2a*change in position
Xf = Xi + Vit + 1/2at^2

Question number 1 is just the velocity formula:
98m/10s = 9.8 m/s

Question number 2 also utilizes the velocity formula:
(250m-100m)/10s = 150m/10s = 15 m/s

Question number 11 uses the acceleration formula:
(15m/s - 25m/s)/2s = (-10m/s)/2s = -5m/s^2 now take the absolute value of that to get magnitude

Question number 12 uses the same concept as number 11. Assign a positive or negative sign respectively to the proper direction.

Number 3-10 use the kinematic formulas I mentioned. I’ll do number 3 for you. The mass of the car plays no factor in the distance traveled. We are given the initial velocity of the car, the acceleration of the car, and the final velocity of the car. It says it came to a rest, so that means 0m/s. The only time independent equation is the Vf squared one. Simply plug in the values and solve for the change in position.
0^2 = 20^2 + 2(-5)*(change in position)
0 = 400 - 10*(change in position)
-400 = -10*(change in position)
Change in position = 40 meters
User Andrea Carron
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