The acceleration of a motorcycle is given by...?

Question

The acceleration of a motorcycle is given by...?

The acceleration of a motorcycle is given by a=At-Br^2 where A=1.20 m/s^3 and B=0.120 m/s^3. It is at rest at the origin at time t=0 s.

a.) What is it's position and velocity as functions of time?
b.) What's the maximum speed it attains?

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Answer 1 · 2018-07-15T19:16:18

Please see the explanation below

Explanation:

The acceleration is

$a (t) = A t - B t^{2}$ $a(t)=At-Bt^2$

The velocity is the integral of the acceleration

$v (t) = \int a (t) d t = \int (A t - B t^{2}) d t$ $v(t)=inta(t)dt=int(At-Bt^2)dt$

$= A \frac{t^{2}}{2} - B \frac{t^{3}}{3} + C$ $=At^2/2-Bt^3/3+C$

Plugging in the initial conditions

$v (0) = 0$ $v(0)=0$

Therefore,

$0 = 0 - 0 + C$ $0=0-0+C$

$C = 0$ $C=0$

So,

The velocity is

$v (t) = \frac{1.2}{2} t^{2} - \frac{0.12}{3} t^{3} = 0.6 t^{2} - 0.04 t^{3}$ $v(t)=1.2/2t^2-0.12/3t^3=0.6t^2-0.04t^3$

The position is the integral of the velocity

$p (t) = \int v (t) d t = \int (0.6 t^{2} - 0.04 t^{3}) d t$ $p(t)=intv(t)dt=int(0.6t^2-0.04t^3)dt$

$p (t) = 0.6 \frac{t^{3}}{3} - 0.04 \frac{t^{4}}{4} + C_{1}$ $p(t)=0.6t^3/3-0.04t^4/4+C_1$

Plugging in the initial conditions

$p (0) = 0 - 0 + C_{1} = 0$ $p(0)=0-0+C_1=0$

Therefore,

The position is

$p (t) = 0.2 t^{3} - 0.01 t^{4}$ $p(t)=0.2t^3-0.01t^4$

The maximum speed is when

$\frac{d v}{d t} = 0$ $(dv)/(dt)=0$

$1.2 t - 0.12 t^{2} = 0$ $1.2t-0.12t^2=0$

$\Rightarrow$ $=>$ , $t (1.2 - 0.12 t) = 0$ $t(1.2-0.12t)=0$

Therefore,

${(t=0),(t=1.2/0.12=10s):}$

The maximum velocity is $v_(max)=60-40=20ms^-1$

See the graph of velocity v/s time

graph{0.6x^2-0.04x^3 [-4.93, 52.82, -2.02, 26.85]}

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The acceleration of a motorcycle is given by...?