**– A rocket initially at rest starts its upward motion from the earth’s surface. The vertical acceleration in +y upward direction in first $10.0s$ of flight is represented by $a_y=(12.8\frac{m}{s^3})t$.**

**– Part(a) – At what elevation will the rocket be at $10.0s$ from the surface of the earth?**

**– Part(b) – When the rocket is $325m$ above the earth’s surface, calculate its speed.**

In this question, we have to find the** height and speed of the rocket** by **integrating** the **acceleration** with the **limits** of time.

The basic concept behind this question is the knowledge of **the kinematics** **equation **of** acceleration, ** **integration, and limits of integration.**

## Expert Answer

Integrate the **kinematics equation** as follows:

\[ v_y=\int_{0}^{t}{a_y}{dt} \]

Now putting the value of $t$ here which is $t=10$:

\[ v_y=\int_{0}^{10}{a_y}{dt}\]

Now putting the value of $a$ here which is given $a=2.8t$:

\[ v_y=\int_{0}^{10}{2.8t}{dt} \]

Now integrating the equation we get:

\[ v_y=2.8(t^ 2)(\dfrac{1}{2})+v_0 \]

Here $v_o$ is the constant which comes after the integration:

\[ v_y = 1.4 t^ 2 + v_0 \]

Here we know that $v_o=0$:

\[ v_y=1.4t^2+(0) \]

\[ v_y=1.4t^2 \]

We also know that:

\[ y=\int_{0}^{10}{v}{dt} \]

Putting $v = 1.4t^2$ in the above equation we get:

\[ y=\int_{0}^{10}{1.4t^2}{dt} \]

Taking derivative we get:

\[ y=1.4(t^3)(\dfrac{1}{3})+y_0 \]

Here we know that $y_0=0$:

\[ y=1.4[ (t^3)(\dfrac{1}{3})]_{0}^{10} + (0) \]

\[ y=\dfrac{1.4}{3}\times [ t^3 ]_{0}^{10} \]

\[ y=0.467 \times [ t^3 ]_{0}^{10} \]

Now substituting the limit of $ t$ in the above equation:

\[ y = 0.467 \times [ (10)^3 – (0)^3 ] \]

\[ y = 0.467 \times [ (10)^3 ] \]

\[ y = 0.467 \times (1000) \]

\[ y = 467 \space m \]

(b) Given we have $ y = 325 \space m $

we know that:

\[ y = \int { v }{ dt } \]

putting $ v = 1.4 t^ 2 $ in the above equation we get:

\[ y = \int { 1.4 t^ 2}{ dt } \]

Taking derivative we get:

\[ y = 1.4 (t^3 ) (\dfrac{1}{3} ) + y_0 \]

here we know that $ y_0 =0 $:

\[ y = 1.4 [ (t^3 ) (\dfrac{1}{3} ) ] + (0) \]

\[ y = 1.4 [ (t^3 \dfrac{1}{3} ) ] \]

\[ y = \dfrac{1.4 }{3} \times [ t^3 ] \]

\[ y = 0.467 \times [ t^3 ] \]

Now substituting the value of $ y $ in the above equation, where $ y = 325 $:

\[ 325 = 0.467 \times [ t^3 ] \]

\[ 325 = 0.467 \times t^3 \]

\[ t =8.86 s \]

Putting it within the limits of the integral we have:

\[ v_y = \int_{0}^{8.86} { 2.8} { dt }\]

\[ v_y = 110 m\]

## Numerical Results

(a) \[y = 467 \space m\]

(b) \[v_y = 110 m\]

## Example

What is the **speed of the rocket** in the above question when it is $300m$ above ground?

We know that:

\[y=0.467 \times [t^3]\]

\[300=0.467 \times [t^3]\]

\[300=0.467 \times t^3\]

\[t=8.57\ s\]

We have:

\[v_y=\int_{0}^{8.57}{2.8}{dt}\]

\[v_y=103\ m\]