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

The time after Jenny’s start that is takes Alyssa to catch up with Jenny is 285 s.

How to calculate the time to catch up?

The time after Jenny’s start that is takes Alyssa to catch up with Jenny is calculated as follows.

Let the speed of Jenny = a

Let the speed of Alyssa = b

Let the time taken to catch up = t

The distance between Jenny and Alyssa when Alyssa started the race is;

D = 3.8 m/s x 15 s

D = 57 m

The time after Jenny’s start that is takes Alyssa to catch up with Jenny is;

(b - a)t = D

(4 m/s - 3.8 m/s)t = 57 m

0.2t = 57

t = 57 / 0.2

t = 285 s

Answer 2

Answer:Answer:

300 seconds

Step-by-step 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 =
`(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=
`(15)/(0.05)`

tJ=300 s

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