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? in problem​

Answer 1

Answer:

2.33m/s

Explanation:

u=0

s1 = 2m, t1 = 1s =>v1 = 2 m/s

s2 = 5m, t2 = 2s=>v2 = 2.5 m/s

Av. Velocity = total distance /total time = 5+2/1+2 = 7/3 =2.33

Av. Acceleration = change in velocity/change in time = (2.33-0)/2

= 1.165 m/s^2

Answer 2

Explanation:-

u = 0 m/s

$s_1 = 2 m \\ t_1 = 1 s \\ s_2 = 5 m \\ t_2 = 3 s$

To find :-

The average velocity and average acceleration.

Solution:-

Since,

Object travel in same direction so displacement covered= Distance travelled.

Total displacement =

→$S = s_1+s_2$

→$S = 5 + 2$

→$S = 7 m$

Total time taken :-

→$T = t_1 + t_2$

→$T = 1 + 2$

→$T = 3s$

Now,

$A_v = \dfrac{\text{Total displacement travelled}}{\text{Toal time taken}}$

→$A_v = \dfrac{7}{3}$

→$A_v = 2.33 m/s$

hence,

Average velocity of object will be 2.33 m/s.

Average acceleration of the object is :-

→$Acc_{av} = \dfrac{u +v}{2}$

→$Acc_{av} = \dfrac{0 + 2.33}{2}$

→$Acc_{av} = \dfrac{2.33}{2}$

→$Acc_{av}= 1.165 m/s^2$

hence,

Average acceleration will be 1.165 m/s².