The x and y co-ordinates of a particle at any time tare given by; x = 7t+4t2 and y = 5t where x and y are in m and t in s. The acceleration of the particle at 5 s is:

vx=dxdt=7+8t⇒ax=dvxdt=8m/s2vy=dydt=5⇒ay=0at t=5s, a=ax2+ay2=82+02=8m/s2

The x and y co-ordinates of a particle at any time tare given by; x = 7t+4t2 and y = 5t where x and y are in m and t in s. The acceleration of the particle at 5 s is:

vx=dxdt=7+8t⇒ax=dvxdt=8m/s2vy=dydt=5⇒ay=0at t=5s, a=ax2+ay2=82+02=8m/s2

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The x and y co-ordinates of a particle at any time tare given by; x = 7t+4t² and y = 5t where x and y are in m and t in s. The acceleration of the particle at 5 s is:

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zero

8 m/s²

20 m/s²

40 m/s²

answer is B.

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$v_{x} = \frac{dx}{dt} = 7 + 8 t \Rightarrow a_{x} = \frac{{dv}_{x}}{dt} = 8 m / s^{2}$
$v_{y} = \frac{d y}{dt} = 5 \Rightarrow a_{y} = 0$
$at t =5 s, a = \sqrt{a_{x}^{2} + a_{y}^{2}} = \sqrt{8^{2} + 0^{2}} = 8 m / s^{2}$

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The x and y co-ordinates of a particle at any time tare given by; x = 7t+4t2 and y = 5t where x and y are in m and t in s. The acceleration of the particle at 5 s is:

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The x and y co-ordinates of a particle at any time tare given by; x = 7t+4t² and y = 5t where x and y are in m and t in s. The acceleration of the particle at 5 s is: