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Question

The motion of a particle is defined by the equations x=4t2 and y=2t3 in meters, where t is in seconds. Determine the normal and tangential components of the particle’s velocity and acceleration when t=2s


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Solution

Step 1: Given Data:

Here, r=(4t2)i^+(2t3)j^

Now we need to differentiate r with respect to t,

v=dr/dt=(8t)i^+(6t2)j^

When t=2s,v=(8×2)i^+(6×22)j^=16i^+24j^

Then, v=162+242v=256+576=832v=28.8m/s

From v value we can say that velocity is directed tangent to the given path.

Now vn=0 and vt=28.8m/s

Then the velocity v makes an angle θ=tan-1(24/16)=56.31°

Step 2 : Calculate acceleration

Accereration, a=dv/dt=8i^+12j^m/s2

at t=2s,a=8i^+12(2)j^=8i^+24j^a=82+242=25.298

Now here acceleration a makes an angle ϕ=tan-1(24/8)=tan-1(3)=71.56°

α=71.560-56.310=15.25°

an=asinα=25.298sin15.25=6.585m/s2at=acosα=25.298cos15.25=24.404m/s2

Hence the normal and tangential components of the particle’s velocity and acceleration when t=2s are vn=0m/s, vt=28.8m/s and an=6.585, at=24.404.


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