Question

If the object is start is to move Intial Point with velocity
10m/sec and accerleration is 8m/ Sec with a time 3 sec then
Find the total distance Covered by the object?​

Answer 1

Answer:

25/4 meters

Explanation:

use v^2-u^2=2as

Answer 2

Question: If the object is start is to move initial point with velocity 10 m/s and acceleration is 8 m/s sq. with time 3 sec then find the total distance covered by the object?

Provided that:

• Initial velocity = 10 m/s
• Acceleration = 8 m/s²
• Time = 3 seconds

To calculate:

• Total distance

Solution:

• Total distance = 66 m

Using concept:

• Second equation of motion

Using formula:

• ${\small{\underline{\boxed{\pmb{\sf{s \: = ut \: + \dfrac{1}{2} \: at^2}}}}}}$

Required solution:

$:\implies \sf s \: = ut \: + \dfrac{1}{2} \: at^2 \\ \\ :\implies \sf s \: = 10(3) + \dfrac{1}{2} \times 8(3)^{2} \\ \\ :\implies \sf s \: = 30 + \dfrac{1}{2} \times 8(9) \\ \\ :\implies \sf s \: = 30 + \dfrac{1}{2} \times 72 \\ \\ :\implies \sf s \: = 30 + 1 \times 36 \\ \\ :\implies \sf s \: = 30 + 36 \\ \\ :\implies \sf s \: = 66 \: m \\ \\ :\implies \sf Distance \: = 66 \: metres$