-Reduce the sample size so the experiment can be done faster.-Increase the sample size from 6 cups to 12 cups of sand and water.-Use more legible handwriting when recording data.-Use more precise digital thermometers.<span>-Use more precise scales that measure to the hundredth of a gram.</span>
Explanation:
The attached figure shows data for the cart speed, distance and time.
For low fan speed,
Distance, d = 500 cm
Time, t = 7.4 s
Average velocity,

Acceleration,

For medium fan speed,
Distance, d = 500 cm
Time, t = 6.4 s
Average velocity,

Acceleration,

For high fan speed,
Distance, d = 500 cm
Time, t = 5.6 s
Average velocity,

Acceleration,

Hence, this is the required solution.
Answer:

Explanation:
consider the mass of each train car be m
m₁ = m₂ = m₃ = m
speed of the three identical train
u₁ = u₂ = u₃ = 1.8 m/s
m₄ = m u₄ = 4.5 m/s
m₅ = m u₅ = 0 (initial velocity )
final velocity
v₁ = v₂ = v₃ = v₄ = v₅ = v
using conservation of momentum
m₁u₁ + m₂u₂ + m₃u₃ + m₄u₄ + m₅u₅ = m₁v₁ + m₂v₂ + m₃v₃ + m₄v₄ + m₅v₅
m (1.8 + 1.8 + 1.8 +4.5) = 5 m v


a) For the motion of car with uniform velocity we have ,
, where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.
In this case s = 520 m, t = 223 seconds, a =0 
Substituting

The constant velocity of car a = 2.33 m/s
b) We have 
s = 520 m, t = 223 seconds, u =0 m/s
Substituting

Now we have v = u+at, where v is the final velocity
Substituting
v = 0+0.0209*223 = 4.66 m/s
So final velocity of car b = 4.66 m/s
c) Acceleration = 0.0209 