Answer
$(a)\space 1.3\space m/s^{2}$
$(a)\space 0.01\space m/s^{2}$
Work Step by Step
Here we use Newton's second law of motion.
Let's write, $F=ma$ where m- mass, a- acceleration, F- force.
(a) Now apply this equation to the single locomotive.
$\vec F=ma$
Let's plug known values into this equation.
$191\space kN=148\space Mg\times a$
$191\times10^{3}\space kgm/s^{2}= 148\times10^{6}\times (\frac{10^{3}g}{10^{3}})\times a$
$\frac{191}{148}\space m/s^{2}= a$
$1.3\space m/s^{2}=a$
(b) Now apply this equation to the two locomotives.
$\vec F=ma$
Let's plug known values into this equation.
$191\times 10^{3}\space kgm/s^{2}= (148+14.3\times10^{3})\times10^{3}\space Mg\times a$
$\frac{191}{14448}m/s^{2}= a$
$0.01\space m/s^{2}= a$
