Here we will look at a more complex problem for which we will use all three of our primary constant acceleration equations. a motorcyclist heading east through a small iowa town accelerates after he passes a signpost at x=0 marking the city limits. his acceleration is constant: ax=4.0m/s2. at time t=0, he is 5.0 m east of the signpost and has a velocity of v0x=15m/s. (a) find his position and velocity at time t=2.0s. (b) where is the motorcyclist when his velocity is 25 m/s?

Question

Damion Austin

Physics

31 January, 13:31

Here we will look at a more complex problem for which we will use all three of our primary constant acceleration equations. a motorcyclist heading east through a small iowa town accelerates after he passes a signpost at x=0 marking the city limits. his acceleration is constant: ax=4.0m/s2. at time t=0, he is 5.0 m east of the signpost and has a velocity of v0x=15m/s. (a) find his position and velocity at time t=2.0s. (b) where is the motorcyclist when his velocity is 25 m/s?

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Pebbles · Answer 1

In physics, it is already convenient since there are already derivedequations for rectilinear motions with constant acceleration. These equations are all derived from Newton's Laws of Motion.

2ax = vf^2 - v0^2

x = v0*t + 1/2*a*t^2

where

a is the acceleration in m/s^2

vf is the final velocity in meters

v0 is the initial velocity in meters

x is the distance travelled in meters and,

t is the time in seconds

a.) For this case, you'll find x when t=2 s. Using the second equation,

x = v0*t + 1/2*a*t^2

x = 15*2 + 1/2*4*2^2

x = 38 m

b.) For this case, you'll find x when vf is 25 m/s. Using the first equation,

2ax = vf^2 - v0^2

2*4*x = 25^2 - 15^2

x = 50 m