Question

an object at rest starts moving it covers a distance of 2m in one second it covers a further distance of 5 m in two seconds in the same direction what is its average velocity and acceleration

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Answer 1

Answer:

Average velocity = 2.33 m/s and Average acceleration = 1.16 m/s²

Explanation:

An object at rest starts moving it covers a distance of 2m in 1 sec it covers a further distance of 5 m in 2 sec in the same direction.

Here..

• D1 = 2 m
• D2 = 5 m

• t1 = 1 sec
• t2 = 2 sec

Distance travelled by object = Total displacement

Now,

Total Displacement = D1 + D2

⇒ 2 + 5

⇒ 7m

Total time = t1 + t2

⇒ 1 + 2

⇒ 3 sec

Now,

Average velocity = Total displacement travelled by object/Total time taken

Substitute the known values in above formula

⇒ 7/3

⇒ 2.33 m/s

Average acceleration of object = Initial velocity + Final velocity/2

We have ..

• u (Initial velocity) = 0 m/s
• v (Final velocity) = 2.33 m/s

Substitute the known values in above formula

⇒ (0 + 2.33)/2

⇒ 2.33/2

⇒ 1.16 m/s²

Answer 2

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

${\bold{\therefore Avg.\:velocity=2.34\:m/s}}$

${\bold{\therefore Avg.\:acceleration=1.55\:m/s^{2}}}$

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

• In the given question information given about an object at rest starts moving it covers a distance of 2m in one second it covers a further distance of 5 m in two seconds in the same direction

• We have to find average velocity and acceleration.

$\underline \bold{Given :} \\ \implies Initial \: velocty(u) = 0 \\ \\ \implies Distance \: cover \: in \: 1\: sec = 2 \: m \\ \\ \implies Distance \: cover \: in \: 2 \: sec = 5 \: m \\ \\ \underline \bold{To \: Find : } \\ \implies Avg. \: velocity = ? \\ \\ \implies Avg.\:acceleration(a) = ?$

• According to given question :

$\bold{For \: Avg. \: velocity : } \\ \implies Avg. \: velocity = \frac{total \: distance}{ \triangle T} \\ \\ \implies Avg. \: velocity = \frac{2 + 5}{1 + 2} \\ \\ \bold{\implies Avg. \: velocity = 2.34 \: ms} \\ \\ \bold{ For \:Avg.\: acceleration : } \\ \\ \bold{By \: third \: equation \: of \: motion : } \\ \implies s = u \: t + \frac{1}{2} \: {a \: t}^{2} \\ \\ \implies 7 = 0 \times 3 + \frac{1}{2} \times a \times { 3 }^{2} \\ \\ \implies 7 \times 2 = 9 \times a \\ \\ \implies a = \frac{14}{9} \\ \\ \bold{\implies a = 1.55 \: {m/s}^{2} }$