Question

A rocket, initially at rest, steadily gains speed for 4.00s while traveling upwards
70.0m.
What was the rocket's final speed?

Answer 1

Answer:

$35.0\; \rm m \cdot s^{-1}$.

Explanation:

The question states that the rocket gains speed steadily. In other words, the acceleration of this rocket in that four seconds would be constant.

• Let $a$ denote the acceleration of this rocket.
• Let $t$ denote the duration of the acceleration.
• Let $v_{0}$ denote the initial velocity of this rocket.
• Let $x$ denote the displacement of this rocket.

Using information from the question:

• $t = 4.00\; \rm s$,
• $x = 70.0\; \rm m$,
• $v_{0} = 0\; \rm m \cdot s^{-1}$ since the rocket was "initially at rest".

Apply the SUVAT equation $\displaystyle \frac{1}{2}\, a \cdot t^{2} + v_{0}\cdot t = x$ to find $a$, the acceleration of this rocket.

$\displaystyle \frac{1}{2}\, a \times (4.00\; {\rm s})^{2} + 0 = 70.0\; \rm m$.

$\begin{aligned}a &= \frac{70.0\; \rm m}{\displaystyle \frac{1}{2} \times (4.00\; {\rm s})^{2}} = 8.75\; \rm m \cdot s^{-2}\end{aligned}$.

In other words, this rocket accelerated from rest at a constant $a = 8.75\; \rm m \cdot s^{-2}$ for $t = 4.00\; \rm s$. The final velocity of this rocket would be:

$\begin{aligned}& v_{0} + a\cdot t \\ =\; & 0 \; \rm m\cdot s^{-1} + 8.75\; \rm m\cdot s^{-2} \times 4.00\; \rm s \\ =\; & 35.0\; \rm m\cdot s^{-1}\end{aligned}$.