A group of students go by car on a day trip by to a nearby beach. Once in the ca

Question

A group of students go by car on a day trip by to a nearby beach. Once in the car they accelerate northward at 3.00 m/s2 until they reach a speed of 86.4 km/hr. They then cruise in that direction for 10 minutes before slowing down to a stop at 6.00 m/s2. They immediately turn right and accelerate at 2.00 m/s2 till they reach 72 km/hr and travel at that speed for another 15 minutes. They then decelerate at 4 m/s2 (still travelling in the same direction) and when they stop they are at the beach. (10 marks) a. How long did the trip take in total? b. How far did they travel? c. How far away from their starting point was the beach? d. What was their average speed? e. What was their average velocity?

Explanation / Answer

northward:

accelerate at 3m/s2 to 86.4kph = 24m/s,
so time = 24/3 = 8s,
dist = 0.5at2 = 1.5*8*8 = 32m

move at this speed for 10 min = 600s
dist = vt = 24*600 = 14400m

decelerate at 6m/s2 to stop
so t = 24/6 = 4s
dist = ut+(1/2)at2 = 24*4 - (0.5)(6)(4*4) = 96-48 = 48m

so total time = 8+600+4 = 612s
total distance = 32+14400+48 = 14480m


turn right, that is, eastward:

accelerate at 2m/s2 to 72kph = 20m/s,
so time = 20/2 = 10s,
dist = 0.5at2 = 1*10*10 = 100m

move at this speed for 15 min = 900s
dist = vt = 20*900 = 18000m

decelerate at 4m/s2 to stop
so t = 20/4 = 5s
dist = ut+(1/2)at2 = 20*5 - (0.5)(4)(5*5) = 100-50 = 50m

total time = 10+900+5 = 915s
total distance = 100+18000+50 = 18150m

a) trip time = sum of northward time+eastward time
= 900+915 = 1815s
b) how far = sum of distances = 14480+18150 = 32630m
c) how far from start point = (144802 + 181502) = 23218.37 m

d) avg speed = distance/time = 32630/1815 = 17.98 m/s

e) avg vel = displacement/time = 23218.37/1815 = 12.79 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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