Question

A rocket is uniformly accelerated from rest to a speed of 1000^-1 in 2.0 min [a]calculate the average velocity and[b] distance traveled

Answer 1

Correct question :-

A rocket is uniformly accelerated from rest to a speed of 1000 m/s in 2 minutes. Calculate (a) the average velocity and (b) the distance travelled

Explanation :-

Given :

• Initial velocity of the rocket,u = 0 [as it is given that it started from rest]

• Final velocity,v = 1000m/s

• Time taken = 2min = 120s   [1min = 60s,so 2min = 2 × 60 = 120s]

To Find :

• Average velocity = ?

• Distance travelled,s = ?

Solution :

According to second law of equation.

$\sf{}\bf{}s=ut+\dfrac{1}{2}at^2$

But we don’t know acceleration,so let’s find out it’s acceleration,from first equation of motion.

$\sf{}v=u+at$

$\sf{}\implies 1000m/s=0+a\times 2s$

$\sf{}\implies 1000m/s=2s\times a$

$\sf{}\implies 1000m/s=a\times 2s$

$\sf{}\implies \dfrac{1000m/s}{2s}=a$

$\sf{}\therefore a=500m/s^2$

Therefore distance:

$\sf{}\implies s=0\times t+\dfrac{1}{2}\times 500\times 2^2$

$\sf{}\implies s=0\times t+\dfrac{1}{2}\times 500\times 4$

$\sf{}\implies s=\dfrac{1}{1}\times 250\times 4$

$\sf{}\therefore s=1000m$

Therefore,distance travelled is equal to  1000m

We know,

$\sf{}Average\ velocity=\dfrac{Total\ distance\ travelled}{Total\ time\ taken}$

So,

$\sf{}\implies \dfrac{1000m}{120s}$

$\sf{}\therefore 3.8m/s$

Therefore,average velocity is equal to 3.8m/s

Answer 2

GiVeN:

A rocket is uniformly accelerated from rest to a speed of 1000 m/s in 2.0 min.

$\begin{gathered}\longmapsto\:\:\bf\blue{Initial\:velocity\:(u)\:=\:0\:m/s\:} \\ \end{gathered}$

$\begin{gathered}\longmapsto\:\:\bf\orange{Final\:velocity\:(v)\:=\:1000\:m/s\:} \\ \end{gathered}$

$\begin{gathered}\longmapsto\:\:\bf\green{Time\:(t)\:=\:2\:min.\:=\:120\:s} \\ \end{gathered}$

$\begin{gathered} \\ \Large{\bf{\pink{\underline{To\:FiNd\::}}}} \\ \end{gathered}$

⑴ The average velocity.

⑵ The distance travelled by rocket.

$\begin{gathered} \\ \Large{\bf{\purple{\underline{CaLcUlAtIoN\::}}}} \\ \end{gathered}$

$\begin{gathered}\bf\red{We\:know\:that,} \\ \end{gathered}$

$\begin{gathered}\orange\bigstar\:\:{\underline{\green{\boxed{\bf{\blue{v\:=\:u\:+\:a\:t}}}}}} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf{1000\:=\:0\:+\:a\times{120}} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf{120a\:=\:1000\:} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf{a\:=\:\dfrac{1000}{120}\:} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf\pink{a\:=\:8.333\:m/s^2} \\ \end{gathered}$

✅ Now we calculate the distance travelled by the rocket and we use the following equation of motion.

$\begin{gathered}\purple\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S\:=\:u\:t\:+\:\dfrac{1}{2}\:a\:t^2}}}}}} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf{S\:=\:0\times{120}\:+\:\dfrac{1}{2}\times{8.333}\times{(120)^2}\:} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf{S\:=\:0\:+\:4.1665\times{14400}\:} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf{S\:=\:59997.6\:m\:\approx\:60000\:m} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf\blue{S\:=\:60\:km} \\ \end{gathered}$

$\begin{gathered} \\ \Large\bf\pink{Therefore,} \\ \end{gathered}$

⑵ The distance travelled by rocket is 60 km.

✅ To calculate average velocity, we use the following formula.

$\begin{gathered}\green\bigstar\:\:{\underline{\blue{\boxed{\bf{\purple{Average\:velocity \:=\:\dfrac{Total\:Displacement}{Total\:Time}\:}}}}}} \\ \end{gathered}$

$\begin{gathered}\bf\red{Where,} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf{Velocity_{(avg.)}\:=\:\dfrac{60000}{120}\:} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf\green{Velocity_{(avg.)}\:=\:500\:m/s} \\ \end{gathered}$

$\begin{gathered} \\ \Large\bf\orange{Therefore,} \\ \end{gathered}$

⑴ The average velocity is 500 m/s.