Question

a car starts from rest and accelerates uniformly, until it has traveled 15km and acquired a velocity of 30m/s. the car then moves at this constant velocity of 30 m/s for 15 minutes. the driver then applied the brakes and the car stopped after 5 minutes.

find

the acceleration during the last part of the trip: a1=

The acceleration during the last part of the trip: a3=

The total distance traveled during the whole trip: dtot=

Answer 1

(a)

Parameters for first part of trip,

As car start from rest, u = 0

v = 30 m/s

s = 15 km = 15*10^3 m

From kinematic equation,

v^2 - u^2 = 2as

30^2 - 0 = 2*a1*15*10^3

a1 = 0.03 m/s^2

(b)

Parameters for last part of the trip,

u = 30 m/s

v = 0

t = 5 min = 5*60 = 300 s

v = u + at

0 = 30 + a3 * 300

a3 = -0.1 m/s^2

(c)

total distance traveled during the whole trip,

d_total = d1 + d2 + d3

here, d1 = 15 km

d2 = 30*15*60 = 27000 m = 27 km

d3 = v^2 / 2a = 30^2 / 2*0.1

d3 = 4500 m = 4.5 km

d_total = 15 + 27 + 4.5

d_total = 46.5 km