Answer only question 5 please A car undergoes a displacement of 500m due east, f

Question

Answer only question 5 please

A car undergoes a displacement of 500m due east, followed by another displacement due south of 1200m.

4. If the car traverses the first part of the trip at 36.0km/h and then accelerates uniformly during the second part until it reaches 144km/h at the end

5. If the car in problem 4, traveling at 144km/h, started back to its starting point in a staight line and decelerated uniformly to reach it as its velocity goes to zero,

a) Calculate its acceleration for the return trip

b) Calculate its average velocity for the return

c) Calculate the time for the return

d) Calculate the average velocity for all three legs of the trip

Explanation / Answer

5)

a) distance to be travelled, d = sqrt(500^2 + 1200^2)

= 1300 m

here, initial speed, u = 144 km/h = 144*5/18 = 40 m/s

final sped, v = 0

so, apply, a = (v^2 - u^2)/(2*d)

= (0^2 - 40^2)/(2*1300)

= -0.615 m/s^2

b) average velocity, V/-avg = (v + u)/2

= ( 0 + 40)/2

= 20 m/s

c) t = d/V_avg

= 1300/20

= 65 s

d) zero.

because, net displacement is zero.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects