Question

An object moves 50m in first 20 sec, 40m in next 20 sec, and 30 m in next 10 sec. What is its average speed? *​

Answer 1

Given :-

An object moves 50m in first 20 sec, 40m in next 20 sec, and 30 m in next 10 sec.

To Find :-

Average speed

Solution :-

◼ I n C a s e 1

Distance = 50 m

Time = 20 sec

◼ I n C a s e 2

Distance = 40 m

Time = 20 sec

◼ I n C a s e 3

Distance = 30 m

Time = 10 sec

Now,

Average Speed = Total distance/Total time

Average Speed = 50 + 40 + 30/20 + 20 + 10

Average Speed = 120/50

Average Speed = 12/5

Average Speed = 2.4 m/s²

Answer 2

Answer:

$\green{ \boxed {\sf \: average \: speed \: \: \bar{v} = 2.4 \: m {s}^{ - 1} }} \:$

Explanation:

We know that,

$\orange{ \boxed{\sf \: average \: speed \: \: \bar {v} = \frac{total \: distance \: (D)}{total \: time \: (T) } }}$

In the question there is three cases.

Case-1

$\sf \: distance \: \: d_1 = 50 \: m \\ \sf \: time \: \: t_1 = 20 \: s$

Case -2

$\sf \: distance \: \: d_2 = 40 \: m \\ \sf \: time \: \: t_2 = 20 \: s$

Case -3

$\sf \: distance \: \: d_3 = 30 \: m \\ \sf \: time \: \: t_3 = 10 \: s$

$\sf \: so \: \: total \: distance \: \: D = d_1 + d_2 + d_3 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = 50 + 40 + 30 \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 120 \: m \\ \\ \sf \: total \: time \: \: T = t_1 + t_2 + t_3 \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 20 + 20 + 10 \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 50 \: s$

So,

$\sf \: average \: speed \: \: \bar{v} = \frac{D}{T} \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{120}{50} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2.4 \: m {s}^{ - 1} \\ \\ \green{ \boxed {\sf \: average \: speed \: \: \bar{v} = 2.4 \: m {s}^{ - 1} }}$