displacement x of a body having relation with time is x= 3t²-6t+7. find the distance covered in first 2 seconds
Answers
Answer:
First, let us find whether the particle reverses its direction of motion during the specified time.
[math]S = 6 +12t -2t^2[/math]
[math]v = \frac{ds}{dt} = 12–4t[/math] … … … (1)
Equating the velocity [math]v[/math] with zero ([math]12–4t =0[/math]) yields [math]t= 3\,s[/math]. This means particle travels for 3 seconds in forward direction and for the next 2 second in backward direction. In order to find the distance in 5 second, we have to find the magnitude of displacement in first 3 second and then 3 second to 5 second and then add the two.
Displacement in first [math]3\,s[/math]:
[math]s(t=3s) - s(t=0) = (6 + 12\times 3 - 2\times 9) - 6 =18 \,m[/math] (assuming SI Unit)
Displacement during 3 to 5 second:
[math]s(t=5 s) - s(t = 3s) = (6+12\times 5–2\times 25) -(6 + 12\times 3 - 2\times 9) =-8\,m[/math]
Ignoring the negative sign and adding the two displacements we have the total distance [math]18 + 8 = 26\,m[/math].
Alternatively, we can find initial velocity u by putting t = 0 in equation (1). This comes out to be 12 m/s. Thus u = 12 m/s.
Displacement in first 3 s:
[math]s = ut +\frac 12at^2 = 12\times 3 + \frac 12 (-4)\times 9 = 18\,m[/math]
Displacement in next 2s (notice that particle starts its backward journey with [math]u = 0[/math]):
[math]s = ut +\frac 12at^2 = 0 - \frac 12 (-4)\times 4 = -8 \,m[/math]