You drive on Interstate 10 from San Antonio to Houston, half thetime at 51 km/h

Question

You drive on Interstate 10 from San Antonio to Houston, half thetime at 51 km/h and the otherhalf at 70 km/h. On the way backyou travel half the distance at51 km/h and the other half at 70 km/h. (a) What is your average speed from San Antonio toHouston?
km/h

(b) What is your average speed from Houston back to SanAntonio?
km/h

(c) What is your average speed for the entire trip?
km/h

(d) What is your average velocity for the entire trip?
km/h
(a) What is your average speed from San Antonio toHouston?
km/h

(b) What is your average speed from Houston back to SanAntonio?
km/h

(c) What is your average speed for the entire trip?
km/h

(d) What is your average velocity for the entire trip?
km/h

Explanation / Answer

(a) What is your average speed fromSan Antonio to Houston?

Vavg1 = [V1t1 +V2t2] / [t1 +t2]   where ttotal1=[t1 + t2] Vavg1 = [51 x 0.5 + 70x0.5] /[0.5 + 0.5]  


Vavg1 = 60.5 km/h
(b)What is your average speed from Houston back to SanAntonio?

Vavg2 = [V3t3 +V4t4] / [t3 + t4] where ttotal2= [t3 +t4]

t3 = V3/(0.5 S ) ; t4 =V4/(0.5 S )

t3 = 0.5 S /V3   ; t4 =0.5 S /V4

Vavg2 = [V3 0.5 S /V3  + V4 0.5 S /V4] / [0.5 S/V3    + 0.5 S/V4 ]

Vavg2 = [1   + 1] / [1/V3    +  1/V4 ]

Vavg2 =2 [V3 V4 ] /[V3    +  V4 ]

Vavg2 = [ 51x70 ] / [51+ 70]
Vavg2 = 59.0 km/h
(c) What is your average speed for the entiretrip?
Vavg = [Vavg1ttotal1 +Vavg2ttotal2 ] /[ttotal1 +ttotal2] ttotal1 = S/Vavg1    ttotal2=  S/Vavg2    wehave Vavg =[Vavg1  S/Vavg1    +Vavg2  S/Vavg2 ] /[S/Vavg1  +S/Vavg2 ] Vavg = [1 +   1 ] /[1/Vavg1     +   1/Vavg2    ] Vavg= 2[Vavg1 Vavg2    ] / [Vavg1  +   Vavg2  ]

Vavg =2[60.5x  59     ] / [60.5 +   59.0 ]

Vavg = 59.75km/h (d) What is your average velocity forthe entire trip?

Zero (0) since you have returned back to San Antonio and yourdisplacement was 0.

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.