An antelope moving with constant acceleration covers the distancebetween two poi
ID: 1735920 • Letter: A
Question
An antelope moving with constant acceleration covers the distancebetween two points 72.0 m apart in6.90 s. Its speed as it passes the secondpoint is 17.0 m/s.(a) What is its speed atthe first point?1 m/s
(b) What is the acceleration?
Explanation / Answer
Remembering the kinematic equations for constant acceleration,using the particular equation: xf = xi + (1/2)(vxi +vxf)t or final pos = initial pos + (1/2)(speed at init. pos + speed at finalpos) * time Plug in to get: 72 = 0 + (1/2)(v + 17)(6.90) => 72 = ((v/2) + (17/2))(6.90) => 72 = 6.90v + 117.30)/2 => 144 = 6.90v + 117.3 => 26.7 = 6.90v => v = 26.7/6.90 ~= 3.87m/s For b, since we now know the initial velocity, we can use the firstequation: vxf = vxi + axt Where a is the acceleration. Plug in to get: 17 = 3.87 + 6.90a => 13.13 = 6.90a => a = 13.13/6.90 ~= 1.90m/s2