Uniform motion question

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.

Answer 1

The difference in the positions is 35 km.

Subtract $1.5S_2$ in your second equation instead of adding.

Answer 2

You are correct in the first part. If we let the biker's speed = $S_1$ and the skater's speed = $S_2$, then $S_1 = 2S_2 + 15$. Note the distance travelled in $1.5$ hours by the biker is:

$$1.5\times{(2S_2 + 15)}$$

Similarly, the distance travelled in $1.5$ hours by the skater is:

$$1.5\times{S_2}$$

The differnce after $1.5$ hours between the biker and skater is $35$km, i.e.:

$$1.5\times{(2S_2 + 15)}-1.5\times{S_2}=35$$ $$1.5\times{(2S_2 + 15-S_2)}=35$$ $${S_2+15}=\frac{35}{1.5}$$ $${S_2}=\frac{35}{1.5}-15$$ $${S_2}=8\frac{1}{3}$$

Your main error was adding the distances rather than subtracting them. I hope this helps!

Answer 3

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 $$(2v+15)\cdot 1.5-1.5v=35$$ $$1.5(2v+15-v)=35$$ $$v+15=\frac{70}{3}$$ $$v=\frac{25}{3}~\text{km}/\text{hr}.$$