How to determine the maximum speed achieved by a motorcycle that accelerates from a standing start and comes to a complete stop after travelling 300m.

Question

The motorcycle is capable of achieving a top speed of 180 kph from a standing start over a distance of 400m with a time of 13.5 seconds.
in an emergency breaking situation the motorcycle requires a distance of 100m to stop from a speed of 100kph

Answer 1

 

Hi Tony:

First, let's put those velocities into units that will play more nicely with the rest of my quantities:

 

180 km/h * (1 h/3600 s) * (1000 m/ 1 km) = 50 m/s

 

100 km/h --> 27.8 m/s

 

calculate the maximum accelerations that the engine can provide, speeding up, and the brakes can provide, slowing down.  If we assume the motorcycle is moving in the positive direction, then the speeding up acceleration should be positive, and the slowing down one negative.  The speeding up one can be determined by the definition for acceleration:

 

a = (v-v_o)/Δt

 

where

 

v = final velocity

v_o = initial velocity

Δt = time interval

 

So, for speeding up, we have:

 

a_su = (50 m/s - 0 m/s)/(13.5 s) = 3.7 m/s²

 

For slowing down, if we want to find a using the kinematic equations, we need to know at least three other kinematic quantities for the slowing down motion.

 

to find: a

know:  v (= 0, since it comes to a stop at the end),  v_o (= 27.8 m/s), d = 100 m

 

Then use the kinematic equation that connects these four quantities:

 

v² = v_o²+2ad 

 

0² = (27.8 m/s)²+2a_sd(100 m)

 

a_sd = -3.86 m/s²

 

So, now we have the speeding up and slowing down acceleration values.  Now we have to look at our restrictions, that the motorcycle speeds up at acceleration a_su and slows down at acceleration a_sd, and that its total displacement is 300 m.

 

Since it would require 400 m to reach 180 kph, and needs 100 m to stop from 100 kph, we can also reasonably conclude that it does not reach top speed (since it only moves 300 m).  So, in our estimation, it must be speeding up with acceleration a_su for the entire first leg and then immediately switching to a_sd slowing down.  Such instantaneous changes in acceleration are unphysical, but make the math much easier to handle, so we usually use them here.

 

Since the kinematic equations depend on there being only one constant acceleration over the time interval of interest, we have to break our interval into the "speeding up" and the "slowing down" part, and set up a kinematic equation for each, that we can hopefully connect up.

 

For the first leg (speeding up), we have known quantities:

 

a = 3.7 m/s²

v_o = 0 m/s (standing stop)

v₁ = final velocity at end of this leg;  we do not know what this is yet

d = d₁   we do not know the displacement over leg 1, either

 

We can set up a kinematic equation relating these quantities:

 

v₁² = (0 m/s)² + 2(3.7 m/s²)d₁   [Equation 1]

 

Unfortunately, this one equation has two unknowns (v₁ and d₁), so we cannot go much further with this alone.

So let's look at the slowing down leg:

 

v_o = v₂ (the initial velocity of the second leg

d = d₂ (the displacement of the second leg, also unknown at this point)

v = 0 (comes to a stop)

a_sd = -3.86 m/s²

 

And we can make a kinematic equation for this leg (the same one as above, just with v=0 instead of v_o = 0):

 

(0 m/s)² = v₂² + 2(-3.86 m/s²)d₂  [Equation 2]

 

This is a second equation, but it has introduced two more unknowns, so now we have four.  The rule of thumb is that, for every unknown in your system, you need one equation connecting those unknowns, if you want to solve for all of them. So, we need two more equations connecting v₁, v₂, d₁, d₂.  Fortunately, we have those in our constraints.

 

We know the final velocity of the first leg must equal the initial velocity of the second, so:

 

v₁ = v₂

 

And the two displacements must add up to 300 m:

 

d₁+d₂ = 300 m

 

Ultimately, we want to solve for the value of v₁ (or v₂, if you wish).  I think solving for one of the displacements first will be easiest; say, d₁.  Then we can use that to solve for v.

 

From the most recent equation, we can write: d₂ = 300 - d₁

 

And, since v₁ = v₂, we can substitute Equation 1 above into Equation 2, and get:

 

0 = v₁ + 2(-3.86 m/s²)d₂ = (0 + 2(3.7)d₁) + 2(-3.86)(300-d₁)

 

0 = 7.4d₁-(2)(3.86)(300)+(2)(3.86)(d₁) = 7.4d₁ - 2316 + 7.72d₁

0 = 15.12d₁-2316

 

d₁ = 2316/15.12 = 153.2 m

 

So, in the first leg, speeding up, it covers 153.2 m.  Since its speeding up and slowing down accelerations have almost the same magnitude, it would make sense that they would split the motion about 50-50.

 

Then we can use Equation 1 above to solve for the maximum speed at the end of the first leg:

 

v² = 0+2(3.7)(153.2) = 1133.7

 

v = 33.7 m/s  (take the positive square root, since we have the motorcycle moving in the positive direction)

 

This is less than the 50 m/s it would achieve after acceleration for 400 m, so it makes sense at least that far.

 

I hope this helps!  Just let me know if there are other questions about this or similar problems.