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Question

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

dvdt=(αβv0)eβt

Where, 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=(αβv)

Considering
α=2m/s2,β=0.05s1 and

dvdt=1.3 m/s2 at t=3s, the distance (in m) travelled by the vehicle in 35 sec is


  1. 900.96

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Solution

The correct option is A 900.96
dvdt=αβv

dv(αβv)=dt

ln(αβv)β=t+C

at t = 3 sec,
dvdt=αβv=1.3 m/s2

ln(1.3)0.05=3+cc=8.247

ln(20.05×v)=(t8.247)×(0.05)

20.05×v=1.51e0.05t

v=dsdt=4030.2e0.05t

s=40t+604e0.05t+c

at t=0,s=0,

c=604

s=40t+604e0.05t604

at t = 35 sec

s=40×35+604×e0.05×35604

s=900.96 m

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