## Related questions with answers

At launch a rocket ship weighs $4.5$ million pounds.
When it is launched from rest, it takes $8.00$ s to reach $161$ km/h; at
the end of the first $1.00$ min, its speed is $1610$ km/h.

(b) Assuming the acceleration is constant during each time inter-
val (but not necessarily the same in both intervals), what distance
does the rocket travel (i) during the first $8.00$ s and (ii) during the
interval from $8.00$ s to $1.00$ min?

Solution

Verified(b)

This is the analysis model for particle under constant acceleration.

To calculate the distance, we will use the following equation:

$\color{Fuchsia} x_f=x_i+\dfrac{1}{2}(v_{xi}+v_{xf})t$

(i) In this time interval the rocket launched from rest, so $v_{xi}=0$

$x_f= 0+\dfrac{1}{2}(0+44.7) (8)$

$\color{#4257b2} = \boxed{178.8\;\mathrm{m}}$

(ii) Here the time interval is not starting from Zero, so the change in $v_x$ from $44.7\;\mathrm{m/s}$ to $ 447.2;\mathrm{m/s}$ will happen in the time interval $t =60-8=52\;\mathrm{s}$

$x_f= 0+\dfrac{1}{2}(44.7+447.2) (52)$

$\color{#4257b2}=\boxed{1.28\times10^4\;\mathrm{m}}$

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