## IB Maths HL Kinematics

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### IB Maths HL Kinematics

IB Mathematics HL – Calculus, Application of Differentiation, Kinematics

Can someone help me with the following exercise?

“A particle moving in a straight line has its displacement given by the following equation

$s(t)= 4t^3-48t^2+180t+24$

a) find the particle’s velocity and acceleration at time t
b) find the time when the particle come to rest. ”

Thanks
Max

Posts: 0
Joined: Mon Jan 28, 2013 8:06 pm

### Re: IB Maths HL Kinematics

IB Maths HL– Calculus, Application of Differentiation, Kinematics

a) Since the velocity (v) is the first derivative of the displacement function, we have

$v(t)=\frac{ds}{dt}=\frac{d}{dt}(4t^3-48t^2+180t+24)=$

$=12t^2-96t+180$

Since the acceleration (a) is the first derivative of the velocity function found above, we have

$a(t)=\frac{dv}{dt}=\frac{d}{dt}(12t^2-96t+180)=$

$=24t-96$

b) The particle come to rest when velocity equals to zero

$v(t)=12t^2-96t+180=0 \Rightarrow t=3 \or\ t=5$

Hope these help!!
miranda

Posts: 268
Joined: Mon Jan 28, 2013 8:03 pm

### Re: IB Maths HL Kinematics

thanks miranda!
Max

Posts: 0
Joined: Mon Jan 28, 2013 8:06 pm