Question

Problem 3: Consider a particle (shown in blue) traveling along a ramp. The particle starts with an initial at the start of time 1 - seconds, at position 1. The length of the ramp is 5 meters, from position / to position 2. The particle is said to have a constant acceleration a -0.86g, where g-10 m/s. velocity 16 m's, a. Determine the velocity of the particle, as it passes position 2 b. Determine the time taken for the particle to travel from position / to position.2 Please Note: Show by integration how you obtain the general formula for the velocity, time taken etc. Do not use memorized or canned formulae. If you do not show your steps, you will v1 = 16 m/s

Answer 1

We know that the acceleration is given by
a = change in velocity/ change in time = dv/dt
$a = \frac{dv}{dt}$
$dv = adt$
integrating both side
$\int_{u}^{v}dv = \int_{0}^{t}adt$
$v-u = at$
$v=u+ at$-------------(1)
Now we know that
v = dx /dt
$\frac{dx}{dt}=u+ at$
$dx=udt+ at*dt$
On integration we get
$\int_{0}^{s}dx=\int_{0}^{t}udt+ at*dt$
$s=ut + \left ( \frac{1}{2} \right )at^{2}$ -----------------(2)
(1) Initial velocity (u) = 16 m/s
acceleration (a) = 0.86g = 0.86*10 = 8.6 m/s²
Distance covered(s) = 5 m
Now for the final velocity first we will use the equation 2
s = ut +(1/2)at²
5 = 16*t +(1/2)*8.6*t²
On solving we get
t = 0.289 s
Now using the equation 1
V = u +at
V = 16 +8.6*(0.289) = 18.49 m/s
Hence the final speed when it passes position 2 will be 18.485 m/s
(2) We have already calculated the time taken i.e 0.289 s