A bike and a car set off from a set of lights at the same time (t=0s)
The car has initial velocity 3m/s and is accelerating at 0.66666m/s/s (2/3m/s/s)
The bike has initial velocity 5m/s and remains at a constant 5m/s.
After 'X' seconds the car has travelled twice as far from the start point as the bike. What is the value of 'X' ?
u-car = 3 m/s
a-car = 2/3 m/s^2
u-bike = 5 m/s
Distance travelled by car in x seconds
= (u-car)(x) + 1/2 (a-car) x^2
= 3x + 1/2 (2/3)x^2
= 3x + 1/3 x^2
Distance travelled by bike
= (u-bike) (x)
= 5x
Given
Distance travelled by car in x seconds = 2 (Distance travelled by bike)
3x + 1/3 x^2 = 2 * 5x = 10x
1/3 x^2 + 3x - 10x = 0
1/3 x^2 - 7x = 0
Multiplying both sides by 3 and factorizing
x(x - 21) = 0
x = 0 and x = 21
Ignore x = 0 since no distance is travelled in 0 second
Soooooo .......
x = 21 seconds
X=21.21 seconds
if you wont to know how to do this use simultaneous equations
vb , vc = velocity bike, care, t - time 0 - intial, D - distance
vc = vc0 + ac x t
vb = vb0
Dc = vc0 x t + 1/2 x ac x t^2
Db = vb0 x t
after t = T seconds we have Dc = 2 x Db so:
vc0 x T + 1/2 x ac x T^2 = 2 x vb0 x T
fill in and solve
x=17 seconds
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