Answer
(a) Yes, the acceleration of the flee is zero for $t\geq1.3ms$.
(b) Thus, the maximum height reached by the flea in the first $2.5ms$ is $0.25cm$.
(c) $a(0.5)=a(1.0)=1.0\times10^5\frac{cm}{s^2}$ and $a(1.5)=0$
(d) $h(0.5)=8.3\times10^{-3}cm$
$h(1.0)=5.0\times10^{-2}cm$
$h(1.5)=0.11cm$
Work Step by Step
The given speed vs time graph can be approximated by the following sketch .
The acceleration has a constant non-zero value till $t=1.3ms$ and then it is zero for the remaining time.
We know that the accleration is given by the slope of the speed vs time graph and the distance covered is given by the area under the speed vs time graph.
(a) Yes, the acceleration of the flee is zero for $t\geq1.3ms$.
This is because the slope of the curve is zero.
(b) The maximum height reached by the flea in the first $2.5ms$ is aproximated by the sum of the area of the triangular and rectangular part of the sketch.
Height reached in $2.5ms$=Area of the triangle+Area of the rectangle=$\frac{1}{2}\times1.3\times10^{-3}\times133+(2.5-1.3)\times10^{-3}\times133=0.25cm$
Thus, the maximum height reached by the flea in the first $2.5ms$ is $0.25cm$.
(c) acceleration at $0.5ms$=acceleration at $1.0ms$=slope of the triangular part of the curve=$\frac{133-0}{1.3\times10^{-3}}=1.0\times10^5\frac{cm}{s^2}$
acceleration at $1.5ms$=slope of the rectangular part of the curve=$0$
(d) Flea's height at $t=0.5ms$=area under the curve (from $t=0$ to $t=0.5ms)=\frac{1}{2}\times0.5\times10^{-3}\times33=8.3\times10^{-3}cm$
Flea's height at $t=1.0ms$=area under the curve (from $t=0$ to $t=1.0ms)=\frac{1}{2}\times1.0\times10^{-3}\times100=5.0\times10^{-2}cm$
Flea's height at $t=1.5ms$=area under the curve (from $t=0$ to $t=1.3ms)+$ area under the curve (from $t=1.3ms$ to $t=1.5ms)=\frac{1}{2}\times1.3\times10^{-3}\times133+(1.5-1.3)\times10^{-3}\times133=0.11cm$
