Question

3. A model rocket takes 0.05 seconds to speed up from rest to its maximum velocity of 80 m/s.

What is its acceleration during launch?

Answer 1

The acceleration of the rocket during launch is
`1600 ms^-^2.`

Step-by-step explanation:

The rocket works on the principle of Newton’s third law which states that every action has an equal and opposite reaction. The average thrusting force from the momentum is given as

`F= \frac{\Delta p} {\Delta t}`

Here Δp is the change in momentum during the process of launching and Δt is the time taken by the rocket to attain its maximum speed for launching.

We also know that
`p = mv` and
`F = ma` from Newton’s second law of motion.

`mass * acceleration=\frac  {(mass * \Delta velcoity)}  {\Delta time}`

As the mass of the rocket will be constant during launching so Δp will have only variation in velocity. Thus the acceleration during launch of rocket will be

`acceleration = (80)/(0.05)= 1600 ms^-^2`

Thus the launching acceleration is 1600 ms-2.

Answer 2

`1600 (m)/(s^2)`

Step-by-step explanation:

Acceleration is defined as the change in velocity divided by the time it took to produce such change. The formula then reads:

`a = (change-in-velocity)/(time) = (Vf-Vi)/(t)`

Where Vf is the final velocity of the object, (in our case 80 m/s)

Vi is the initial velocity of the object (in our case 0 m/s because the object was at rest)

and t is the time it took to change from the Vi to the Vf (in our case 0.05 seconds.

Therefore we have:

`a = (80 m/s - 0 m/s)/(0.05 sec) = 1600 (m)/(s^2)`

Notice that the units of acceleration in the SI system are
`(m)/(s^2)` (meters divided square seconds)