Question

the position a the rocket sled on a straight track is given as x= at^3+ bt^2+c, where a= 1.04m/s^2, b=1.94 m/s^2, and c= 8.29m. what is the sled's average speed between t= 3.72 and t=6.79

Answer 1

V = x / t

If x = at³ + bt² + c where:

a = 1.04 m/s³ In this case, it should have s³ and not s² because when you are simplyfing the units, it does not match with the other units of b and c, correct me if I'm wrong with the data;

b = 1.94 m/s²; c = 8.29 m

t₁ = 3.72 s; t₂ = 6.79 s

Then to calculate the average speed, you need to calculate x₁ and x₂ and use 1 of 2 methods:

1. Once you have X₁ and x₂, calculate V₁ and V₂ and then, calculate the average;

2. Calculate the average of x₁ and x₂, average of t₁ and t₂ and then, the speed v.

Both methods are fine but I recommend you to use the first method, because you are actually calculating the position in both times, in the second method you are only taking a average of the time, and this is not always really accurate; you may choose whatever you like but I'll do it by the first method:

x₁ = 1.04x(3.72)³ + 1.94x(3.72)² + 8.29 = 88.67 m

v₁ = 88.67 / 3.72 = 23.84 m/s

x₂ = 1.04x(6.79)³ + 1.94x(6.79)² + 8.29 = 423.3 m

v₂ = 423.3 / 6.79 = 62.34 m/s

v = 62.34 + 23.84 / 2 = 43.09 m/s   -----> This should be the average speed.

If you do it by method 2:

x = 423.3 + 88.67 / 2 = 255.985 m -------> t = 3.72 + 6.79 / 2 = 5.255

v = 255.985 / 5.255 = 48.71 m/s -----> 5 m/s of difference between two methods; it's not that significant but still counts at the end.

If you need anything else, you can tell me.