Question #128862

1) A particle is constrained to travel along the path. If x = (4t^2)m, where t is in seconds, determine the magnitude of the particle’s velocity and acceleration when t= 0.5 s.

2) A particle traveling along a parabolic path y = 0.25x^2. If x= (2t^2)m, where t is in seconds,

determine the magnitude of the particle’s velocity and acceleration when t = 2s.

2) A particle traveling along a parabolic path y = 0.25x^2. If x= (2t^2)m, where t is in seconds,

determine the magnitude of the particle’s velocity and acceleration when t = 2s.

Expert's answer

(i)

Given"\\ x = 4t^2"

as we know that velocity is,V = "\\frac{dx}{dt}"

so "\\frac{dx}{dt} = 8t"

velocity when t = 0.5 ; V = "8\\times0.5 = 4 m\/s"

Acceleration is "\\frac{d^2x}{dt^2}" which is equal to 8. Hence acceleration is 8 m/s^{2}

(ii) Given "y = 0.25x^2 \\ and \\ x= 2t^2"

substituting value of x in y we get, "y = 0.25 \\times[2t^2]^2"

"y= t^4"

Velocity (V) is "\\frac{dy}{dt}" and acceleration (a) is "\\frac{d^2y}{dt^2}"

"and \\ \\frac{dy}{dt} = 4t^3"

so velocity at t= 2sec; V = "4\\times 2^3 = 32m\/s"

and "\\frac {d^2y}{dt^2}= 12t^2"

so acceleration at t= 2 sec ;

a = "12\\times 2^2 = 48m\/s^2"

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