Question

an object is a started from west with a constant acceleration of 3m/s² for 10s. then find out final velocity after 10s also distance covered by object in 10s.

Do this by equation of motion ​

Answer 1

Answer:

v=u+at

v= 0 + 10×3

v= 30m/s

S= ut+1/2 at^2

S= 0×10+ 1/2 × 3× 10×10

S= 150m

Answer 2

Solution :-

As per the given data ,

• Initial velocity (u) = 0 m/s
• Acceleration (a) = 3 m/s ²
• Time taken (t) = 10s

As the object is moving with uniform acceleration we can use the first equation of motion in order to find time(t) and second equation of motion in order to find distance (s)

By applying the first equation of motion ,

➜ v = u + at

Now let's substitute the given values in the above equation ,

➜ v = 0 + 3 x 10

➜ v = 30 m/s

By using the second equation of motion ,

➜ s = ut + at² / 2

➜ s = at²/2 (∵u = 0 )

➜ s = 3 x 100 / 2

➜ s = 3 x 50

➜ s = 150 m

The distance covered by the object in 10s is 150 m