Uniform motion question A biker and a skater set out at 4:30 PM from the same point, headed in the same direction. The biker is travelling at a rate of 15 km/hr faster than twice the speed of the skater. In 1.5 hours, the biker is 35 km ahead of the skater. Find the rate of the skater.
My approach:
Biker speed = $S_1$, Skater speed = $S_2$
$S_1 = 2S_2 + 15$
$1.5(2S_2 + 15) + 1.5S_2 = 35$
I get $2.8 \ \text{km}/\text{hr}$ for $S_2$
Which is not correct. Tell me where I made the mistake.

The equation of motion for the biker is $x_1(t)=(2v+15) t$. The equation of motion of the skater is $x_2(t)=vt.$ We know that $x_1(1.5)-x_2(1.5)=35.$ Thus
$$(2v+15)\cdot 1.5-1.5v=35$$
$$1.5(2v+15-v)=35$$
$$v+15=\frac{70}{3}$$
$$v=\frac{25}{3}~\text{km}/\text{hr}.$$
