Part A n antelope moving with constant acceleration covers h

Part A n antelope moving with constant acceleration covers he distance 77.0 m between two points in time 7.50 s ts speed as it passes the second point is 14.9 m/s What is its speed at the first point? m/s Submit My Answers Give Up Part B What is the acceleration? m/s2 Submit My Answers Give Up

Solution

Problem 8

s = (1/2)(u + v)t

I assume your first point is the initial velocity. s = 77, v = 14.9 and t = 7.5

77 = (1/2)(u + 14.9)(7.5)

5.633 = u

The speed at the first point is 5.633 m/s

Question 2
Use the formula, v = u + at where a is the acceleration.

14.9 = 5.633 + a(7.5)

1.235 = a

The acceleration is 1.235 m/s²

Problem 9 :

Let initial velocity = vo,
velocity after first 4.7s = v1,
velocity after second 4.7s = v2

vo = 0
v1 = vo + 4.7 a = 0 + 4.7 a
v1 = 4.7 a------------(1)
v2 = v1 + 4.7 a---------(2)
Substituting the value of v1 from (1) into (2),

v2 = 9.4 a------------(3)

Distance traveled in the second 4.7s = [(v1+v2)/2] * 4.7
Therefore [(v1+v2)/2] * 4.7 = 200
v1 + v2 = 85.10------------(4)
Substituting the values of v1 and v2 from (1) and (3) into (4),
4.7a + 9.4a = 85.10

a = 6.03 m/s^2
Distance traveled in the first 5.0s
= vo*4.7 + (1/2) * a* 4.7^2
= 0 + (1/2) * 6.03 * 4.7*4.7
= 66.60
Ans: 66.60 m