Answer
a) 13.1 m/s
b) 0.94 $m/s^{2}$
Work Step by Step
Let the car's velocity = b
$distance = speed \times time$
At 14s,
$d=(b)(14)=14b$
At 28s,
$d=(b)(28)=28b$
Let train's acceleration = a
$x=ut+\frac{1}{2}at^{2}$
At 14s,
$x=0+\frac{1}{2}a(14)^{2}=98a$
At 28s,
$x=0+\frac{1}{2}a(28)^{2}=392a$
At 14s, the car has traveled the same distance as the train + length of train
$14b=98a-92.......equation\;1$
At 28s, the car has traveled the same distance as the train
$28b=392a...........equation\; 2$
We now use a system of linear equations:
$28b=196a-184........equation\; 1 \times$2
Sub in equation 2:
$(392a)=196a-184$
$196a=184$
$a=\frac{46}{49}\approx0.94\; m/s^{2}$
$28b=392a$
$28b=392(\frac{46}{49})$
$b=\frac{92}{7}\approx13.1$ m/s
