Question

) A skater with mass m=68.5 kg moving initially at 2.4 m/s on rough horizontal ice, comes to

rest uniformly in 3.52 s due to the friction from the ice. What force does friction exert on

the skater? [4 mks]

b) The acceleration of a particle is given by a(t)=At – Bt2

, with A=1.5 m/s3

and B=1 m/s4

. The

particle is at rest at time t=0 at the origin.

i) Find its velocity as function of time [3 mks]

ii) Calculate vmax [3 mks]

c) Write the expression of the centripetal acceleration in an uniform circular motion in terms of

the period T, the time for one revolution

Answer 1

Explanations & Calculations

a)

• With the use of the equation $\small s=\large{\frac{(v+u)}{2}}\small t$ you can calculate the distance he travels before coming to a stop. Here $\small v =0$ as he comes to a stop finally.
• Once you have the travelling distance, using energy conservation under non-conservative forces can be used to calculate the applied force of friction.

$\qquad\qquad \begin{aligned} \small \frac{1}{2}mu^2&=\small fs\\ \small f&=\small \frac{mu^2}{2s} \end{aligned}$

b)

i)

• Upon integration of the given function with respect to time, you get the function meant for speed as a function of time.

$\qquad\qquad \begin{aligned} \small v(t)&=\small \int a(t) dt\\ &=\small \int At-Bt^2\\ &=\small A\int t dt-B\int t^2dt\\ &=\small A\frac{t^2}{2}-B\frac{t^3}{3} +k\end{aligned}$

• k is a constant and it may be found using the conditions given.
• The particle was at rest when $\small t=0$ meaning the speed was zero at that time. You may use this to find k.

ii)

• To find the max v, you need to differentiate this speed function and consider the maxima and minima.
• But you again get the acceleration function upon differentiation, hence you can directly equal the acceleration function to zero and get the corresponding time.
• And that time can be substituted in the speed function to get the maximum speed.

c)

• It is given by ,

$\qquad\qquad \begin{aligned} \small a&=\small r\omega^2\\ &=\small r\Big(\frac{2\pi}{T}\Big)^2\\ &=\small \frac{4\pi^2 r}{T^2} \end{aligned}$