Constant acceleration calculations.

[cary1234]

Homework Statement

an antelope moving with constant acceleration covers the distance between two points 70.0 m apart in 7.0 s. Its speed as it passes the second point in 15.0 m/s .

a) what is its speed at the first point?
b)what is the acceleration?

Homework Equations

what do you mean by relevant equations??
please give me a little more idea.

The Attempt at a Solution

well, honestly I am advancing to what my teacher teach on us. because she is absent in the last meeting. :( i figure that X sub 0 is 70.0 m and I am stuck at that. i can't figured out how to determine the symbol that i will use.

i think i can use the formula x = X sub 0 + V sub 0x + 1/2 asubx t^2

 

[tiny-tim]

Homework Equations

what do you mean by relevant equations??
please give me a little more idea.

Write out all the standard constant acceleration equations.

In part 3, choose one of them, and use it to solve the problem … it will have to include v_i v_f s and a.

 

[kerimek]

but i don't know how to determine V sub 0x and asubx

I can provide a response to this content by first addressing the student's confusion about relevant equations. Relevant equations in physics are equations that directly relate to the problem at hand and can be used to solve it. In this case, the relevant equations would be those related to constant acceleration, such as the equation for displacement (x = x0 + v0t + 1/2at^2) and the equation for final velocity (v = v0 + at).

Now, to solve this problem, we can use the given information to find the missing variables. We know that the displacement (x) is 70.0 m, the time (t) is 7.0 s, and the final velocity (v) is 15.0 m/s. We can plug these values into the equation for final velocity to find the initial velocity (v0):

v = v0 + at
15.0 m/s = v0 + a(7.0 s)
Solving for v0, we get v0 = 0.71 m/s

Now, to find the acceleration (a), we can use the equation for displacement:

x = x0 + v0t + 1/2at^2
70.0 m = 0 + (0.71 m/s)(7.0 s) + 1/2a(7.0 s)^2
Solving for a, we get a = 2.71 m/s^2

Therefore, the speed at the first point (a) is 0.71 m/s and the acceleration (a) is 2.71 m/s^2.

In summary, by using the relevant equations for constant acceleration, we were able to solve for the missing variables and provide a response to the given content. This is a common approach in science, where we use equations and data to analyze and understand phenomena.