Question

A particle is moving on a straight line with speed
v = (30t^2 + 40t + 2) (where v is in m/s and t is in
s). The average speed from t= 0 to t = 5 s​

Answer 1

Answer:

$\displaystyle{V_{ave}=352 \ m/sec}$

Explanation:

Given :

Velocity ( v ) = 30 t² + 40 t + 2

We have to find average velocity :

We know :

$\displaystyle{V_{ave}=\dfrac{Total \ displacement}{time}}$

$\displaystyle{V_{ave}=\dfrac{\int\limits^5_0 {30t^2+40t+2} \, dt }{5}}\\\\\\\displaystyle{V_{ave}=\dfrac{\left[30t^3/3+40t^2/2+2t\right]^5_0}{5}}$

$\displaystyle{V_{ave}=\dfrac{\left[10t^3+20t^2+2t\right]^5_0}{5}}}\\\\\\\displaystyle{V_{ave}=\dfrac{10\times125+20\times25+10}{5}}\\\\\\\displaystyle{V_{ave}=\dfrac{1250+500+10}{6}}\\\\\\\displaystyle{V_{ave}=\dfrac{1760}{5} \ m/sec}\\\\\\\displaystyle{V_{ave}=352 \ m/sec}$

Hence we get average speed .

Answer 2

Answer:

${\bold{\therefore V_{Avg}=352\:ms^{-1}}}$

Explanation:

${\bold{\huge{\underline{SOLUTION-}}}}$

• In the given question information given about a particle moving on a straight line and time limit is given.

• We have to fins the average speed of particle.

$\underline \bold{Given : } \\ \implies v = ({30t}^{2} + 40t + 2)m {s}^{ - 1} \\ \\ \underline \bold{to \: find : } \\ \implies Average \: Speed = ?$

• According to given question :

$\bold{Basic \: things : } \\ \\ \bold{Differentiation} \\ \implies Distance \implies Velocity \implies Acceleration \\ \\ \bold{Integration} \\ \implies Accleration \implies Velocity \implies Distance \\ \\ \implies v = {30t}^{2} + 40t + 2 \\ \implies \frac{dx}{dt} = 30 {t}^{2} + 20t + 2 \\ \implies dx = (30 {t}^{2} + 20t + 2)dt \\ \bold{Intergrating \: both \: side}\\ \int dx = \int (30 {t}^{2} + 20t + 2)dt \\ \implies x = \frac{30 {t}^{3} }{3} + \frac{20 {t}^{2} }{2} + 2t \\ \implies x = 10 {t}^{3} + 20 {t}^{2} + 2t \\ \bold{For \: average \: velocity} \\ \implies v_{avg} = \frac{Total \: displacement}{Total \: time} \\ \implies v_{avg} = \frac{ 1 0 {t}^{3} + 20 {t}^{2} + 2t }{5} \\ \bold{Putting \: limit \: 0 \to \:5} \\ \implies v_{avg} = \frac{10 \times {5}^{3} + 20 \times {5}^{2} + 2 \times 5}{5} \\ \implies v_{avg} = \frac{1250 + 500 + 10}{5} \\ \implies v_{avg} = \frac{1760}{5} \\ \bold{\implies v_{avg} = 352 \: m {s}^{ - 1} }$