A car initially at rest travels for 2 minutes with a uniform acceleration of 0.3 m/s^2,after which the speed is kept constant until the car is brought to rest with a uniform retardation of 0.6 m/s^2.If the total distance covered is 4500m,what is the time taken for the journey?

Answer 1

First leg of trip:
a = .3
v = .3t + c
when t = 0 , v = 0, so c = 0
v = .3t
s = .15t^2 + k
when t = 0, s = 0, so k = 0

so s = .15t^2 is the distance after t seconds
when t = 2 min = 120sec
v = .3(120) = 36 m/s
s = .15(120^2) m = 2160 m
so the car went 2160 m before hitting the brakes, and had a velocity of
36 m/s
Middle part of trip:
constant speed of 36 m/s for t seconds
distance covered = 36t

3rd leg of trip
a = -.6
v = -.6t + c
so initially t = 0 again, v = .3(120) = 36 m/s <---- velocity from first trip
36 = c
v = -.6t + 36
s = - .3t^2 + 36t + k
when t = 0, distance covered by 2nd trip was 0
k = 0

s = -.3t^2 + 36t

we want the car to stop, so v = 0
-.6t + 36 = 0
t = 60
so it takes 60 seconds to stop, and the distance covered in that time
= -.3(60^2) + 36(60)
= -1080 + 2160 = 1080 m

total distance = 2160 + 36t + 1080 = 4500
36t = 4500-2160-1080 = 1260
t = 35 seconds

total time = 120+35+60 seconds = 215 seconds or 3 minutes and 35 seconds

Answer 2

Write any two different between mks and CGS system