The acceleration- time relationship for a vehicle subjected to non- un

1.	The acceleration- time relationship for a vehicle subjected to non- uniform acceleration isdvdt=(α−βv0)e−βtWhere, v is the speed in m/s, t is the time in s, α and β are parameters, and v0 is the initial speed in m/s. If the accelerating behaviour of a vehicle, whose driver intends to overtake a slow moving vehicle ahead, is described as,dvdt=(α−βv)Consideringα=2m/s2,β=0.05s−1 anddvdt=1.3 m/s2 at t=3s, the distance (in m) travelled by the vehicle in 35 sec is900.96
Answer» The acceleration- time relationship for a vehicle subjected to non- uniform acceleration is $\frac{d v}{d t} = (α - β v_{0}) e^{- β t}$ Where, v is the speed in m/s, t is the time in s, $α$ and $β$ are parameters, and $v_{0}$ is the initial speed in m/s. If the accelerating behaviour of a vehicle, whose driver intends to overtake a slow moving vehicle ahead, is described as, $\frac{d v}{d t} = (α - β v)$ Considering $α = 2 m / s^{2}, β = 0.05 s^{- 1} and$ $\frac{d v}{d t} = 1.3 m / s^{2} a t t = 3 s$ , the distance (in m) travelled by the vehicle in 35 sec is 900.96

The acceleration- time relationship for a vehicle subjected to non- uniform acceleration isdvdt=(α−βv0)e−βtWhere, v is the speed in m/s, t is the time in s, α and β are parameters, and v0 is the initial speed in m/s. If the accelerating behaviour of a vehicle, whose driver intends to overtake a slow moving vehicle ahead, is described as,dvdt=(α−βv)Consideringα=2m/s2,β=0.05s−1 anddvdt=1.3 m/s2 at t=3s, the distance (in m) travelled by the vehicle in 35 sec is900.96

Answer»

The acceleration- time relationship for a vehicle subjected to non- uniform acceleration is

$\frac{d v}{d t} = (α - β v_{0}) e^{- β t}$

Where, v is the speed in m/s, t is the time in s, $α$ and $β$ are parameters, and $v_{0}$ is the initial speed in m/s. If the accelerating behaviour of a vehicle, whose driver intends to overtake a slow moving vehicle ahead, is described as,

$\frac{d v}{d t} = (α - β v)$

Considering

$α = 2 m / s^{2}, β = 0.05 s^{- 1} and$

$\frac{d v}{d t} = 1.3 m / s^{2} a t t = 3 s$ , the distance (in m) travelled by the vehicle in 35 sec is

900.96

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