The displacement of a body at time t seconds is x = −9−9 t^3−9 t^2 m. Find velocity and acceleration when t=4?
The long winded wording: The displacement of a body from the origin O, at time t seconds is x = −9−9 t^3−9 t^2 metres. Find the velocity and acceleration when t = 4.
I need to know how to get the velocity in m/s and the acceleration in m/s^2
Pleeeease help me with this
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Verified answer
x(t) = −9−9 t^2−9 t^3
so
v(t) = dx/dt (t) = -18t - 27t² (in m / s)
and
a(t) = dv/dt (t) = - 18 - 54t (in m/s²)
hope it' ll help !!
"Solution: Given x(t) = â9â9 t^2â9 t^3
velocity =v(t) =
dx/dt (t) = -18t - 27t² m / s
at t=4
velocity =v(4) = dx/dt (4) = -18*4 -
27*4² m / s
velocity =v(4)=-504
means in the negative direction
and
acceleration=a(t) = dv/dt (t) = - 18 –
54t m/s²
at t=4
acceleration=a(4) = dv/dt (4) = - 18 –
54t*4m/s²
acceleration=a(4)=-234
means in the negative direction "