Question

Alan leaves Los Angeles at 8:00 a.m. to drive to San Francisco, 400 mi away. He travels at a steady 50 mph. Beth leaves Los Angeles at 9:00 a.m. and drives a steady 60 mph. a. Who gets to San Francisco first? b. How long does the first to arrive have to wait for the second?

Short answer

(a) Person B will first rich Place S.

(b) Person B will wait for 20 minutes for Person A to arrive.

Step-by-step solution

Given information

The velocity of person A $v_A=50mph$

The velocity of person B $v_B=60mph$

Distance to place S $x=400mi$

Part (a)Time required for person A to reach place S.

Person A is moving with velocity$50mph$.

The time required for person A to cover 400 miles is

localid="1647961767711" $$\begin{gathered} t_A=\frac{x}{v_A} \\ {t}_A=\frac{400}{50} \\ {t}_A=8hours \end{gathered}$$

Person A takes $8hours$to reach the place S.

A person A will reach place S at $8hours$from $8:00am$, that is, at $4:00pm$.

Time required for Person B to reach Place S.

Person B is moving with velocity $60mph$.

The time required for Person B to cover 400 miles is

localid="1647962584022" $$\begin{gathered} t_B=\frac{x}{v_B} \\ {t}_B=\frac{400}{60} \\ {t}_B=\frac{20}{3}hours \\ =6hours40minutes \end{gathered}$$

Person B takes localid="1647962614155" $6hours40minutes$to reach Place S.

A Person A will reach Place S at $6hours40minutes$from $9:00am$, that is, at $3:40pm$.

Therefore, Person B will first reach place S.

Part (b)

Person A will reach Place S at $4:00pm$and Person B will reach place S at $3:40pm$.

A person A has to wait for Person B for

localid="1648451136316" $4:00pm-3:40pm=20minutes$

Therefore, Person B has to wait for 20 minutes for person A.