Answer to Question #311253 in Mechanics | Relativity for Honey Jones Gucor

Question

In a 300-m race, runner A starts from rests and accelerates at 1.6m/s2

for the first 30 m and then runs

at a constant speed. Runner B starts from rests and accelerates at 2.0m/s2

for the first 30m and then

runs at a constant speed. Runner A begins running as soon as the race begins but B firsts takes a nap

to rest up. What is the longest nap that B can take and still not to lose the race?

Answer 1

Solution.

$S=300m;$

$v_{0A}=0;$

$a_A=1.6 m/s^2;$

$v_{0B}=0;$

$a_B=2.0 m/s^2;$

$S_A=S_B=30m;$

$S_a=\dfrac{v_A^2}{2a_A}\implies v_A=\sqrt{2a_AS_A};$

$v_A=\sqrt{2\sdot1.6\sdot30}=9.8 m/s;$

$a_A=\dfrac{{v_A-v_{0A}}}{t_A}\implies t_A=\dfrac{v_A-v_{0A}}{a_A};$

$t_A=\dfrac{9.8}{1.6}=6.1s;$

$S_B=\dfrac{v_B^2}{2a_B}\implies v_B=\sqrt{2a_BS_B};$

$v_B=\sqrt{2\sdot2.0\sdot30}=11.0 m/s;$

$a_B=\dfrac{{v_B-v_{0B}}}{t_B}\implies t_B=\dfrac{v_B-v_{0B}}{a_B};$

$t_B=\dfrac{11.0}{2.0}=5.5s;$

$\tau_A=\dfrac{270}{9.8}=27.6 s;$

$\tau_B=\dfrac{270}{11.0}=24.5 s;$

$\Delta t=(t_A+\tau_A)-(t_B+\tau_B);$

$\Delta t=(6.1+27.6)-(5.5+24.5)=3.7s;$

Answer: $\Delta t=3.7s.$