Question

5.
A particle starting from rest moves with constant acceleration if it takes 5.0 sec to reach the
speed 18 km/hr find (1) average speed during this period (2) distance travelled by the particle
during this period​

Answer 1

Answer:

average speed = 2.5 m/s

distance travelled = 12.5 m

Explanation:

particle start from rest so its initial velocity is zero

now by using the equation, v = u + at

we can find the acceleration = 1 m/s^2

now using the equation, S = ut + 1/2 at^2

we can find the distance travelled = 12.5 m

now we know that,

average speed = total distance/total time taken

so average speed = 12.5/5  = 2.5 m/s

Answer 2

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

$\green{\therefore{\text{ Avg.\:velocity=2.5\:m/s}}}$

$\green{\therefore{\text{ Distance=12.5\:m}}}$

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

• 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.

$\green{\underline \bold{Given :}} \\ :\implies \text{Initial \: velocity(u) = 0} \\ \\ :\implies \text{Final \: velocity(v) = 18 \:km/h} \\ \\ :\implies \text{Time(T) = 5 \: sec} \\ \\ \red{\underline \bold{To \: Find : }} \\ :\implies Avg \: velocity = ? \\ \\ :\implies \text{Distance \:travelled(s) = ?}$

• According to given question :

$\bold{v = \cancel{18} \times \frac{5}{ \cancel{18}} = 5 \: m/s }\\ \\ \bold{For \: Avg \: velocity : } \\ \\ \bold{By \:first \: equation \: of \: motion : } \\ :\implies {v} = {u} +at \\ \\ :\implies 5 = 0 + a \times 5 \\ \\ :\implies a = \frac{ \cancel{5}}{ \cancel5} \\ \\ \green{:\implies a = 1 \: m/ {s}^{2} } \\ \\ \bold{By \: third \: equation \: of \: motion : } \\ :\implies s = ut + \frac{1}{2} a {t} \\ \\ :\implies \frac{s}{t} = 0 + \frac{1}{ \cancel2} \times \cancel{1 }\times 5 \\ \\ :\implies Avg \: velocity = 2.5 \\ \\ \green{:\implies avg \: velocity = 2.5\: m/s} \\ \\ \bold{For \: distance : } \\ \\ \bold{By \: second \: equation \: of \: motion : } \\ :\implies {v}^{2} = {u}^{2} + 2as \\ \\ :\implies {5}^{2} = {0}^{2} + 2 \times 1 \times s \\ \\ :\implies 25 = 2 \times s \\ \\ :\implies s = \frac{ \cancel{25}}{ \cancel{2}} \\ \\ \green{:\implies s = 12.5 \: m}$