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 accelerationof the cart from the speed vs. time data: Use any tool you wish(Excel, MathCAD, calculator). [Of course, you can directly use theequations above, but you would probably find that a bittedious!]
answer in m/s
b.) What is the best estimate of the mass of the cart? inkilograms.
The students add 0.180 N tothe force applied to the cart.
c.) What will the cartâs acceleration be now? in m/s/s.
They now put an additional 0.257 kg mass on the cart (they still havethe extra 0.180 N appliedforce).
d.) What will the cartâs acceleration be now?
4m/s2
e.) How much mass must they now add to the cart to make thecart's acceleration what it was initially? [Note: A negative addedmass means that they must remove mass from the cart.]
5 kg
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 accelerationof the cart from the speed vs. time data: Use any tool you wish(Excel, MathCAD, calculator). [Of course, you can directly use theequations above, but you would probably find that a bittedious!]
answer in m/s
b.) What is the best estimate of the mass of the cart? inkilograms.
The students add 0.180 N tothe force applied to the cart.
c.) What will the cartâs acceleration be now? in m/s/s.
They now put an additional 0.257 kg mass on the cart (they still havethe extra 0.180 N appliedforce).
d.) What will the cartâs acceleration be now?
4m/s2
e.) How much mass must they now add to the cart to make thecart's acceleration what it was initially? [Note: A negative addedmass means that they must remove mass from the cart.]
5 kg
