Question

An obiect travels 15 m in 8 s and then another in
25 m 12 s. What is the average speed of the object
a) 2 ms-1 b) 4 ms-1 c) 4 ms 1 d) 2 ms1​

Answer 1

$\large\underline{\underline{\sf Given:}}$

• Distance $\sf{(s_1)}$ = 15m

• Time $\sf{(t_1)}$ = 8s

• Distance $\sf{(s_2)}$ =25m

• Time $\sf{(t_2)}$ = 12s

$\large\underline{\underline{\sf To\:Find:}}$

• Average Speed $\sf{(V_{av}}$ = ?

$\large\underline{\underline{\sf Solution:}}$

$\large{\boxed{\sf Average\: Speed=\frac{Distance}{Time} }}$

$\large\implies{\sf V_{av}=\frac{s_1+s_2}{t_1+t_2}}$

$\large\implies{\sf V_{av}=\frac{15+25}{8+12}}$

$\large\implies{\sf V_{av}=\frac{40}{20}}$

$\large\implies{\sf V_{av}=2\:m/s }$

$\Large\underline{\underline{\sf Answer:}}$

•°•

$\red{\boxed{\sf Average\: Speed (V_{av})=2\:m/s}}$

Answer 2

Explanation:

$\huge\underline{\underline{\sf \: Answer}}$

$\underline{Given \: that = > }$

Distance in 8 sec = 15m

Distance in another 12 sec = 25m

$average \: speed = \frac{total \: distance}{total \: time}$

$average \: speed = \frac{15 + 25}{8 + 12}$

$average \: speed = \frac{40}{20} = 2m {s}^{ - 1}$

$(a)2m {s}^{ - 1} \: is \: correct$