displacement-velocity relationdisplacement-time relationconstant accelerationgravityinitial velocityfinal velocitykinematic equation1-D motion

Motion Along a Straright Line

  • Straight-line motion, average and instantaneous x-velocity:

vav=ΔxΔt=x2−x1t2−t1v=lim⁡Δt→0ΔxΔt=dxdt\begin{aligned} v_{av} &= \frac{\Delta x}{\Delta t} = \frac{x_2-x_1}{t_2-t_1}\\ v &= \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt} \end{aligned}

  • Average and instantaneous x-acceleration:

aav=ΔvΔt=v2−x1v2−t1a=lim⁡Δt→0ΔvΔt=dvdt\begin{aligned} a_{av} &= \frac{\Delta v}{\Delta t} = \frac{v_2-x_1}{v_2-t_1}\\ a &= \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = \frac{dv}{dt} \end{aligned}

  • Straight-line motion with constant acceleration
    When the x-acceleration is constant, four equations relate the position xx and the x-velocity vv at any time tt to the initial position x0x_0, the initial x-velocity v0v _{0} (both measured at time t=0t = 0), and the x-acceleration aa.

x=v0+vf2⋅tvf=v0+a⋅tx=v0t+12at2vf2=v02+2a⋅x\begin{aligned} x &= \frac{v_0 + v_f}{2} \cdot t \qquad &v_f &= v_0 + a\cdot t\\ x &= v_0 t + \frac{1}{2}at^2 \qquad &v_f^2 &= v_0^2 + 2a\cdot x \end{aligned}

Exercises

1. Describe the motion at interval [t0,t6][t_0, t_6].
Graph 1

Solution

[t0,t2][t_0, t_2], Moving backward at a constant acceleration.
[t2,t4][t_2, t_4], Moving forward at a constant acceleration.
[t4,t6][t_4, t_6], At rest.
[t6,)[t_6, ), Moving forward at a constant accleration.

2. Grasshopper jumps upward, and reaches a max height of 1m1m. Find the total time it is in the air. (up and down).
Graph 2

Solution

When the grasshopper reaches the max height, vf=0m/sv_f = 0m/s.

vf2=v02+2aya=9.8m/s2,y=1m0=v02+2⋅9.8⋅1⇒v0=19.6m/s\begin{aligned} v_f^2 &= v_0^2 + 2ay & a=9.8m/s^2, y=1m\\ 0 &= v_0^2 + 2\cdot9.8\cdot 1 \To v_0 = \sqrt {19.6} m/s \end{aligned}

When the grasshopper finally hits the ground, yf=0my_f = 0m

yf=v0t+12at20=19.6t+12⋅9.8⋅t2⇒t≈0.89s\begin{aligned} y_f &= v_0 t + \frac{1}{2}at^2 \\ 0 &= \sqrt {19.6} t + \frac{1}{2} \cdot 9.8 \cdot t^2\\ \To t &\approx 0.89s \end{aligned}

3. An object is thrown upward on a cliff 20m20m from the ground at an intial velocity of 30m/s30m/s. Find the final velocity before it hits the ground.
b. Find how long is the object in the air.
Graph 3

Solution

a. According the kinematic equation, vf2=v02+2ayv_f^2 = v_0^2 + 2ay. We have v0=−30m/s,a=9.8m/s2,y=20mv_0=-30m/s, a=9.8m/s^2, y=20m (moving downward being positive direction).

vf2=(−30)2+2⋅9.8⋅20⇒vf=1292m/s≈35.94m/s\begin{aligned} v_f^2 &= (-30)^2 + 2\cdot 9.8 \cdot 20\\ \To &v_f = \sqrt {1292} m/s \approx 35.94m/s \end{aligned}

(Mid Term 1) A stone is thrown outward from the top of a 59.4-m high cliff with an upward velocity component of 19.5m/s. How long is the stone in the air.
Graph

Solution

yf−y0=v0t+12at2yf=0my0=59.4ma=−9.8m/s2v0=19.5m/s0−59.4=19.5t−12⋅9.8t2⇒t=6s\begin{aligned} y_f - y_0 &= v_0 t + \frac{1}{2} a t^2\\ y_f &= 0m\\ y_0 &= 59.4m\\ a &= -9.8m/s^2\\ v_0 &= 19.5m/s\\ 0-59.4 &= 19.5t - \frac{1}{2} \cdot 9.8 t^2\\ \To t &= 6s \end{aligned}