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

`35.0\; \rm m \cdot s^(-1)`.

Step-by-step 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 (1)/(2)\, a \cdot t^(2) + v_(0)\cdot t = x` to find
`a`, the acceleration of this rocket.

`\displaystyle (1)/(2)\, a * (4.00\; {\rm s})^(2) + 0 = 70.0\; \rm m`.

`\begin{aligned}a &= \frac{70.0\; \rm m}{\displaystyle (1)/(2) * (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) * 4.00\; \rm s \\ =\; & 35.0\; \rm m\cdot s^(-1)\end{aligned}`.