Motion: An antelope moving with constant acceleration covers the distance between two

[Jeffster]

I have been trying to figure out a practice problem from my textbook all day and can't figure it out. The problem reads: "An antelope moving with constant acceleration covers the distance between two points 66.0m apart in 6.70s. Its speed as it passes the second point is 14.4m/s. What are (a) its speed at the first point and (b) it acceleration."

The useful data that I've been able to get out of it are:
X(0)= 0m
X(6.70s) = 66.0m
V_av = 9.851m/s
V(0) = question (a)
V(6.70s) = 14.4m/s
a_av = 2.149m/s²
a = question (b)

I can't find any equation in the book that explicitly tells me how to get the answers from what I have, and any simultaneous solution attempts have winded up in misery.

 

[stapel]

I have been trying to figure out a practice problem from my textbook all day and can't figure it out. The problem reads: "An antelope moving with constant acceleration covers the distance between two points 66.0m apart in 6.70s. Its speed as it passes the second point is 14.4m/s. What are (a) its speed at the first point and (b) it acceleration."

The useful data that I've been able to get out of it are:
X(0)= 0m
X(6.70s) = 66.0m
V_av = 9.851m/s
V(0) = question (a)
V(6.70s) = 14.4m/s
a_av = 2.149m/s²
a = question (b)

I can't find any equation in the book that explicitly tells me how to get the answers from what I have, and any simultaneous solution attempts have winded up in misery.

What relationships have they given you between position, velocity, and acceleration? What have you tried so far? Even if it "wound up in misery", we'll need to see your efforts.

Please be complete. Thank you!

 

[Jeffster]

There's 4 equations that have been highlighted in the book:
_Vx = =V_0x + a_xt
X = X₀ + V_0xt + (1/2)(a_xt²)
V_x² = V_0x² + 2a_x(x - x₀)
X - X₀ = (1/2)(V_0x + V_x)t

There's also some equations that the book used to help explain the reasoning behind what's going on, and then combined them with others and manipulated them to get the 4 above, so I'm not too sure how useful these will be but here they are:
a_x = (V_2x - V_1x)/(t₂ - t₁)
V_av-x = (X - X₀)/t
V_av-x = (1/2)(V_0x + V_x)
V_av-x = V_0x + (1/2)a_xt

Unfortunately my attempts so far have either been scribbled out so that I can't read it anymore, or has had the page torn out. I am also rather tired right now and can't make another attempt to show how far I get. First thing in the morning I'll do it and post the results.

 

[Deleted member 4993]

I have been trying to figure out a practice problem from my textbook all day and can't figure it out. The problem reads: "An antelope moving with constant acceleration covers the distance between two points 66.0m apart in 6.70s. Its speed as it passes the second point is 14.4m/s. What are (a) its speed at the first point and (b) it acceleration."

The useful data that I've been able to get out of it are:
X(0)= 0m
X(6.70s) = 66.0m
V_av = 9.851m/s
V(0) = question (a)
V(6.70s) = 14.4m/s
a_av = 2.149m/s²
a = question (b)

I can't find any equation in the book that explicitly tells me how to get the answers from what I have, and any simultaneous solution attempts have winded up in misery.

Assume that the antelope is travelling in straight line (linear motion).
Let the

time elapsed = t = 6.7: distance traveled = s = 66; final speed = v = 14.4 ;initial speed = u ; acceleration = a

We have two unknowns - we need two equations.

v = u + at → 14.4 = u + 6.7 a ...............................................(1)

s = ut + 1/2 * at^2 → 66 = 6.7 u + 1/2(6.7)^2*a → 9.850746 = u + 3.35*a ......................................(2)

Two equations & two unknowns and continue.....u = 5.30149

and continue

 

[Jeffster]

Okay, so I just noticed that the equation X - X₀ = (1/2)(V₀ + V)t had all but one known variable. so at t =6.17s it is: 66.0m - 0m = (1/2)(V₀ + 14.4m/s)6.17
and I managed to get V₀ = 5.301m/s from it.
It was the subscripts that were confusing me. Never really had to use them before and they made me forget that X is the same as X(t).