Question

A particle moves in an XY-plane in such a way that its x and y-coordinates vary with time according to x(t)=t3-32t, y(t)=5t2+12.find the acceleration of the parricle ,if t=3s

Answer 1

For acceleration we are required to find Velocity in x axis as well as y axis by differentiation.

After finding velocity we will again differentiate it with respect to time.

Vx = dx/dt

Vx = d(t³-32t)/dt

Vx = 3t²-32

Similarly,

Vy = dy/dt

Vy = d(5t²+12)/dt

Vy = 10t

Again Differentiating w.r.t 't'......

ax = dVx/dt

ax = 6t

if t = 3s

ax = 18

Now,

ay = dVy/dt

ay = 10

Again,

|a| = √(ax)²+(ay)²

|a| = √(18)²+(10)²

|a| = 20.5 ms-² Answer......

Answer 2

Answer:

Ar = 18(i) + 10(j) m/sec^2 [Vector form]

|Ar| = 20.59 m/sec^2 [Magnitude of acceleration]

Explanation:

Provided that:

particle moves in an XY-plane in such a way that it's x-coordinates vary with time as f(t);

$x(t)= {t}^{3} -32t \: \: ...equation(i)$

Differentiating equation (i) twice we get the acceleration of particle on X-axis, Ax;

$x(t)= {t}^{3} -32t\\ v = \frac{dx}{dt} = 3 {t}^{2} - 32 \\ a = \frac{ {d}^{2} x}{d {t}^{2} } = 6t \\ |a| (at \: t = 3) = 18 \: m. {sec}^{ - 2}$

So we get Ax = 18 m/sec^2

And y-coordinates vary with time according to f(t);

$y(t)=5 {t}^{2} +12 \: \: ...equation(ii)$

Differentiating equation (ii) twice we get the acceleration of particle on Y-axis, Ay;

$y(t)=5 {t}^{2} +12 \\ v = \frac{dy}{dt} = 10 t \\ a = \frac{ {d}^{2} x}{d {t}^{2} } = 10 \: (time \: independent) \\ |a| (at \: t = 3) = 10 \: m. {sec}^{ - 2}$

So we get Ay = 10 m/sec^2

Hence the resultant acceleration on XY plane;

Ar = Ax(i) + Ay(j)

(i) and (j) are unit vector towards X and Y axis respectively;

$Ar = \sqrt{ {Ax}^{2} + {Ay}^{2} } \\ Ar = \sqrt{ {18}^{2} + {10}^{2} } \: m. {sec}^{ - 2} \\ Ar = 20.59 \: m. {sec}^{ - 2}$

So the acceleration of the particle ;

Ar = 18(i) + 10(j) unit [Vector form]

|Ar| = 20.59 m/sec^2 [Magnitude of acceleration]

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