Question

Jenny and Alyssa are members of the cross-country team. On a training run, Jenny starts off and runs at a con-stant 3.8 m/s. Alyssa starts 15 s later and runs at a constant 4.0 m/s. At what time after Jenny’s start does Alyssa catch up with Jenny?

Answer 1

Answer:Answer:

300 seconds

Explanation:

We solve this problem using the equation of uniform motion;

d  =v*t

d: Distance traveled

t:  Time

v: Speed

We raise the equations of uniform motion because they run at a constant speed.

dJ=3.8*tJ   Motion equation for Jenny

dA=4*tA   Motion equation for Alyssa

d:distancia

When Jenny and Alissa meet, they will have traveled the same distance,then:

dJ=dA

3.8 tJ  =4*tA

tA =$\frac{3.8}{4}$ * tJ

tA =0.95¨*tJ Equation (1)

Jenny runs 15 seconds longer than Alyssa,then:

tJ = tA+15 Equation (2)

We replace tA =0.95tJ of the Equation (1) in the Equation (2):

tJ = 0.95*tJ +15

tJ-0.95*tJ=15

0.05* tJ=15

tJ=$\frac{15}{0.05}$

tJ=300 s

Alyssa catch up with Jenny in 300 s after Jenny’s start .