v[m/s] data .158, .273, .465, .409, .529, .543, .720, .785 t[s] data .5, 1, 1.5,

Question

v[m/s] data .158, .273, .465, .409, .529, .543, .720, .785

t[s] data .5, 1, 1.5, 2, 2.5, 3, 3.5, 4

a.) Use linear least squares to find the acceleration of the cart from the speed vs. time data: Use any tool you wish (Excel, MathCAD, calculator). [Of course, you can directly use the equations above, but you would probably find that a bit tedious!]
answer in m/s

b.) What is the best estimate of the mass of the cart? in kilograms.

The students add 0.180 N to the force applied to the cart.

c.) What will the cart’s acceleration be now? in m/s/s.

They now put an additional 0.257 kg mass on the cart (they still have the extra 0.180 N applied force).

d.) What will the cart’s acceleration be now?
4 m/s2

e.) How much mass must they now add to the cart to make the cart's acceleration what it was initially? [Note: A negative added mass means that they must remove mass from the cart.]
5 kg

Explanation / Answer

without equipment, you can use the linear regression's equation, so just simply plug 2 pair of values into the equation to have the acceleration. a=(v1-v2)/(t1-v2)=0.18(m/s2)

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