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Wize University Dynamics Textbook (Master) > Kinematics: Rectilinear Motion

Uniformly Accelerated Motion

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If begin with the basic dynamics equations:

v=x˙=dxdtv = \dot{x}=\frac{dx}{dt}

a=v˙=dvdt=x¨=d2xdt2=vdvdxa = \dot{v}=\frac{dv}{dt}=\ddot{x} = \frac{d^2x}{dt^2}=v\frac{dv}{dx}

If we assume the acceleration term (a) to be a constant, we can integrate the various forms of these equations to arrive at the following:

v=vo+atv= v_o+at

x=xo+vot+1/2at2x = x_o + v_ot+1/2 at^2

v2−v02=2a(x−x0)v^2-v_0^2 = 2a(x-x_0)

The important thing to remember is that these 3 equations can only be used under conditions of uniform acceleration. Common applications of this include projectile motion (due to the earth's gravity), as well as motion at a constant velocity (a = 0).

Application of these equations is fairly straightforward, as long as you can keep track of initial conditions at t=0 vs. other conditions at various times. Some key ideas to remember:
  • The maximum displacement occurs when the velocity is 0
  • If the system is starting from rest, the initial velocity is 0
  • We often don't consider xo if we are looking for distance or displacement - xo is just a reference point
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You throw a ball vertically with a velocity of 20 m/s upwards from the second floor of the engineering building at a height of 7 m from the ground. Determine:

a) The maximum height that the ball reaches, and how long does that take
b) The velocity of the ball as its about to hit the ground
c) The time it takes for the ball to reach the ground


a)vo=20, xo=7, a=−9.81v_o=20,\ x_o=7,\ a=-9.81

max⁡ h @ v=0\max\ h\ @\ v=0

v2−vo2=2adv^2-v_o^2=2ad

d=(20)22(9.8)=20.4→h=20.4+7=27.4md=\frac{\left(20\right)^2}{2\left(9.8\right)}=20.4\to h=20.4+7=27.4m

vf=vo+atv_f=v_o+at

0=20−9.81t→t=2.04s0=20-9.81t\to t=2.04s

b) vf=?, x=0v_f=?,\ x=0

v2−vo2=2a(x−xo)v^2-v_o^2=2a\left(x-x_o\right)

v2=(20)2+2(−9.81)(0−7)→v=23.18 msv^2=\left(20\right)^2+2\left(-9.81\right)\left(0-7\right)\to v=23.18\ \frac{m}{s}

vf=vo+atv_f=v_o+at

−23.18=20−9.81t→t=4.40s-23.18=20-9.81t\to t=4.40s


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While checking your email on your phone as you go up in a helicopter accelerating upwards at a a rate of 2 m/s2, you drop your phone. At that instant the helicopter (and your phone) are travelling upwards at a velocity of 40 m/s. After 14 seconds, your phone signal goes out (indicating that it has hit the ground). Determine:

Note: do not assume that the helicopter began accelerating from the ground. The acceleration of gravity is 9.81 m/s2 downwards.

a) At what speed did your phone hit the ground

b) what is the height of the helicopter.
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Mark Yourself Question
  1. Grab a piece of paper and try this problem yourself.
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You've timed the traffic light near your house and found that it stays red for exactly 30 seconds. One day you're driving down the road and see the light turn red as you are 3 blocks (300 meters) away from the intersection, and your speedometer reads 60 km/h. Determine the constant rate of deceleration of the car you need in order to not stop at the traffic light. What is your speed as you enter the intersection?