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A particle moves along the path y^2= 4x where x and y are in metres. The x co-ordinate of the particle at any time is given by x=t^2 The acceleration at x = 4m is ms 1. 1 2. 2.5 3. 2 4. 3 ?
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A particle moves along the path y^2= 4x where x and y are in metres. T...
Solution:

Given data: y^2= 4x, x=t^2

Let's differentiate x w.r.t t, then

dx/dt= 2t

Now, let's substitute the value of x in the equation of y^2= 4x, then

y^2= 4(t^2)

y= 2t

Differentiating y w.r.t t, we get

dy/dt= 2

Now, let's find the acceleration (a) at x= 4m, then

x= 4m, t= 2s

dx/dt= 2t= 2(2)= 4m/s

Substituting the value of t in y= 2t, we get

y= 2(2)= 4m

Differentiating y w.r.t t, we get

dy/dt= 2m/s

Now, using the chain rule, we get

a= (d^2x/dt^2)= (d/dt)(dx/dt)= (d/dt)(4)= 0

Therefore, the acceleration at x= 4m is 0 ms^-1.

Answer: 1. 0
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A particle moves along the path y^2= 4x where x and y are in metres. The x co-ordinate of the particle at any time is given by x=t^2 The acceleration at x = 4m is ms 1. 1 2. 2.5 3. 2 4. 3 ?
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