Question

The x and y coordinates of a particle at any time at any time are x = 5t - 3t² and y = 4t² respectively, where x and y are in metres and t in seconds. Find the magnitude of acceleration of the particle at t=2s.



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Answer 1

Answer:

$\boxed{\mathfrak{Magnitude \ of \ accelration \ (a) = 10 \ m/s^2}}$

Explanation:

x and y coordinates\ Position of a particle at any time is given as:

x = 5t - 3t² m

y = 4t² m

Rate of change in position is known as velocity.

So, velocity in x and y axis is:

$\rm v_x = \frac{dx}{dt} \\ \\ \rm v_x = \frac{d(5t - 3 {t}^{2} )}{dt} \\ \\ \rm \implies v_x = 5 - 6t \: m {s}^{ - 1} \\ \\ \rm v_y = \frac{dy}{dt} \\ \\ \rm v_y = \frac{d(4 {t}^{2} )}{dt} \\ \\ \rm \implies v_y = 8t \: m {s}^{ - 1}$

Rate of change in velocity is known as accelration.

So, accelration in x and y axis is:

$\rm a_x = \frac{dv_x}{dt} \\ \\ \rm a_x = \frac{d(5 - 6 t)}{dt} \\ \\ \rm \implies a_x = - 6 \: m {s}^{ - 2} \\ \\ \rm a_y = \frac{dv_y}{dt} \\ \\ \rm a_y = \frac{d(8t )}{dt} \\ \\ \rm \implies a_y = 8 \: m {s}^{ - 2}$

$\rm Accelration \: ( \vec{a}) = -6 \hat{i} + 8 \hat{j}$

Magnitude of acceleration of particle (a):

$\rm \implies | \vec{a}| = \sqrt{ {a_x}^{2} + {a_y}^{2} } \\ \\ \rm \implies a = \sqrt{ {( - 6)}^{2} + {8}^{2} } \\ \\ \rm \implies a = \sqrt{36 + 64} \\ \\ \rm \implies a = \sqrt{100} \\ \\ \rm \implies a = 10 \: m {s}^{ - 2}$