An antelope moving with constant acceleration covers the distance between two points 70.0
m apart in 7.00 s. Its speed as it passes the second point is 15.0 m/s. (a) What is its speed at the first
point? (b) What is its acceleration?
d = Vi t + .5 a t^2
70 = 7 Vi + 24.5 a
v = Vi + a t
15 = Vi + 7 a
so Vi = (15 - 7 a)
70 = 7(15-7a) + 24.5 a
solve for a
go back and solve for Vi
ERER
5 m/s
To find the speed at the first point, we can use the equation of motion:
v^2 = u^2 + 2aS
where:
v = final velocity
u = initial velocity
a = acceleration
S = distance
Since the antelope covers the distance between the two points in 7.00 s, we can deduce that the final velocity at the second point (v) is 15.0 m/s.
(a) What is its speed at the first point?
To find the speed at the first point, we need to solve for the initial velocity (u). Rearranging the equation, we have:
u^2 = v^2 - 2aS
Substituting the given values into the equation:
u^2 = (15.0 m/s)^2 - 2a(70.0 m)
To find the value of u, we need to know the acceleration (a). Let's solve for the acceleration first.
(b) What is its acceleration?
To find the acceleration, we can use the equation of motion:
v = u + at
Since the antelope is moving with constant acceleration, we can assume that the final velocity (v) is equal to the speed at the second point (15.0 m/s). So, we can rewrite the equation as:
15.0 m/s = u + a * 7.00 s
To solve for the acceleration (a), we need another equation. We know that the antelope covered a distance of 70.0 m in 7.00 s. We can use the equation of motion that relates distance, initial velocity, acceleration, and time:
S = ut + (1/2)at^2
Substituting the given values into the equation:
70.0 m = u(7.00 s) + (1/2)a(7.00 s)^2
Now, we have a system of two equations with two unknowns (u and a). We can solve this system of equations simultaneously using algebra, substitution, or elimination to find the values of u and a.
After determining the values of u and a, you can calculate the speed at the first point using the equation:
u = initial velocity
Note: The values and units used in the explanation are hypothetical. Please use the actual values from the problem to solve it.