Question

A particle starting from rest moves with constant acceleration if it takes 5.0s to reach the speed 180 km/h find the average velocity during this period and the distance travelled by the particle during this period

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Avg.\:velocity=25\:m/s}}$

${\bold{\therefore Distance=125\:m}}$

${\bold{\underline{\underline{Step-by-Step\:explanation:}}}}$

• In the given question information given about a particle start moving from rest and after 5 sec velocity increase 180 km/h.

• We have to find the avg velocity and distance travelled by the particle during this time period.

$\underline \bold{Given :} \\ \implies Initial \: velocity(u) = 0 \\ \\ \implies Final \: velocity(v) = 180 \:kmh \\ \\ \implies Time(T) = 5 \: sec \\ \\ \underline \bold{To \: Find : } \\ \implies Avg \: velocity = ? \\ \\ \implies Distance \:travelled(s) = ?$

• According to given question :

$\bold{v = \cancel{180} \times \frac{5}{ \cancel{18}} = 50 \: m/s }\\ \\ \bold{For \: Avg \: velocity : } \\ \\ \bold{By \:first \: equation \: of \: motion : } \\ \implies {v} = {u} +at \\ \\ \implies 50 = 0 + a \times 5 \\ \\ \implies a = \frac{ \cancel{50}}{ \cancel5} \\ \\ \bold{\implies a = 10 \: m/ {s}^{2} } \\ \\ \bold{By \: third \: equation \: of \: motion : } \\ \implies s = ut + \frac{1}{2} a {t} \\ \\ \implies \frac{s}{t} = 0 + \frac{1}{ \cancel2} \times \cancel{10 }\times 5 \\ \\ \implies Avg \: velocity = 25 \\ \\ \bold{\implies avg \: velocity = 25\: m/s} \\ \\ \bold{For \: distance : } \\ \\ \bold{By \: second \: equation \: of \: motion : } \\ \implies {v}^{2} = {u}^{2} + 2as \\ \\ \implies {50}^{2} = {0}^{2} + 2 \times 10 \times s \\ \\ \implies 2500 = 20 \times s \\ \\ \implies s = \frac{ \cancel{2500}}{ \cancel{20}} \\ \\ \bold{\implies s = 125 \: m}$

Answer 2

Solution:

Given:

➜ A particle starting from rest moves with constant acceleration if it takes 5.0s to reach the speed 180 km/h.

Find:

➜ Find the average velocity during this period and the distance travelled by the particle during this period.

Know terms:

➜ Initial velocity = (u)

➜ Final velocity = (v)

➜ Acceleation = (a)

➜ Time = (T)

Given:

➜ v = 180 × 5/18 = 50 m/s

Average velocity:

Using first equation of motion:

➜ v = u + at

➜ 50 = 0 + a × 5

➜ a = 50/5

➜ a = 10 m/s²

Using third equation of motion:

➜ s = ut + 1/2 at

➜ s/t = 0 + 1/2 × 10 × 5

➜ 25 m/s

Distance:

Using second equation of motion:

➜ v² = u² + 2as

➜ 50² = 0² + 2 × 10 × 5

➜ 2500 = 20 × 5

➜ s = 2500/20

➜ s = 125 m

#AnswerWithQuality

#BAL