Question

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.

Answer 1

(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/s²

(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 = $12\times 2^2 = 48m/s^2$