Answer to Question #94130 in Mechanics | Relativity for Saundy

Question

Lean and m who are 150m apart start walking toward each other at 3.0m/s and 5.0m/s respectively. How long will it take for them to meet? Find the corresponding distance travelled by each.

Answer 1

Given:

$d = 150\text{ m}; v_1 = 3.0\frac{\text{m}}{\text{s}}; v_2 = 5.0\frac{\text{m}}{\text{s}}.$

Relative speed of one object to the other, since they are moving towards each other, will be (imagine it as speed of one object in the reference frame of the other being stationery therein):

$v = v_1 + v_2.$

Time passed until meeting (all distance covered):

$t = \frac{d}{v} = \frac{d}{v_1+v_2} = \frac{150\text{ m}}{3.0\frac{\text{m}}{\text{s}} + 5.0\frac{\text{m}}{\text{s}}} = 18.75\text{ s}.$

Distances traveled by each object respectively:

$d_1 = v_1t = \frac{v_1}{v_1+v_2}d = \frac{3.0\frac{\text{m}}{\text{s}}}{3.0\frac{\text{m}}{\text{s}} + 5.0\frac{\text{m}}{\text{s}}}150\text{ m} = 56.25\text{ m}; \\ d_2 = v_2t = \frac{v_2}{v_1+v_2}d = \frac{5.0\frac{\text{m}}{\text{s}}}{3.0\frac{\text{m}}{\text{s}} + 5.0\frac{\text{m}}{\text{s}}}150\text{ m} = 93.75\text{ m}.$