On turning a corner, a motorcyclist rushing at 44 ms-1 finds a child on the road 100m away. He applies

Question

On turning a corner, a motorcyclist rushing at 44 ms-1 finds a child on
the road 100m away. He applies the brakes so as to stop the motorcar
within with in 1 m of the child. Calculate the time required to stop.

Answer 1

Hello Jennifer:

 

In kinematics, there are five kinematic quantities:  displacement (x-x_o), initial velocity (v_o), final velocity (v), acceleration (a), and time (t).  There are usually four kinematic equations relating these five quantities:

 

1.  x-x_o = v_o(t)+1/2(a)(t²)     (does not involve v)

2.  x-x_o = (1/2)(v+v_o)t           (does not involve a)

3.  v=v_-o+at                          (does not involve (x-x_o))

4.  v²=v_o²+2a(x-x_o)              (does not involve t)

 

Note that each equation involves four of the kinematic quantities, so, to solve for one, you need to know three others.  In this case, it breaks down like this:

 

to find:  t

know: x-x_o, v_o, v

 

x-x_o  = 99 m (assuming that the cycle comes to a stop exactly 1 m from the child)

vo = 44 m/s

v= 0 m/s (as the cycle has come to a stop)

 

So, we need to choose the equation that involves (x-x_o), v_o, v, and t, but not a, which we have not considered.  You could calculate a, as the  answer below suggests, and then use one of the equations that involve a.  But there is an equation above that involves only those four quantities directly:  Equation 2.  Using that, we get:

 

99 m = (1/2)(44 m/s+0 m/s)(t)

 

99 = (1/2)(44)t = 22t

t = 99/22 = 4.5 s

 

in one step (as opposed to two, as shown below; there is nothing wrong with two steps, but it is not necessary in this case)

 

With those kinematic equations above, as long as you know any three of the quantities, you can calculate a fourth.  The one caveat is that you always either have to either know or be trying to calculate v_o, as that appears in every equation (it is a very important quantity for determining the later motion of an object).

 

 

Answer 2

Hi Jennifer,

 

You know:

Δx = 99m, the displacement that the motorcycle will travel before coming to a stop, which is 1 m from the child who is initially 100 m away.

v₀ = 44 m/s, the initial velocity

v = 0, the final velocity

t = ?, the unknown time to stop

a = ?, the acceleration is unknown

 

You can apply the following kinematic equation:

 

v² = v₀² + 2aΔx

 

v = 0, so

a = -v₀² / 2Δx

 

a = -(44 m/s)² / 2(99 m) = -9.78 m/s². The is approximately the acceleration due to gravity.

 

Now that we know the acceleration we can use another kinematic formula to solve for the time required to stop.

 

v = v₀ + at

 

But, v = 0, so

 

t = -v₀ / a

 

or

 

t = -(44 m/s) / (-9.78 m/s²)

 

t = 4.49 s, the time required to stop the motorcycle