Question

if object A moves with velocity 10m/s and object B moves with acceleration 0.5m/s^2 at what point they meet

Answer 1

GIVEN:

Initial velocity = 50 m/s.

Acceleration of the object is 10 m/s².

Time taken = 16 s

TO FIND:

Distance covered by the object = 2080 m

EXPLANATION:

$\boxed{\bold{\large{S = ut + \dfrac{1}{2}at^2}}}$

u = 50 m/s

t = 16 s

a = 10 m/s²

S = 50(16) +1/2 10(16²)

S = 800 + 5 × 256

S = 800 + 1280

S = 2080 m

∴ The distance covered by the object will be equal to 2080 m.

Some more formulae:

$\boxed{\bold{\large{v = u + at}}}$

$\boxed{\bold{\large{v^2 - u^2 =2 as}}}$

Answer 2

Answer:

Initial velocity = 50 m/s.

Acceleration of the object is 10 m/s².

Time taken = 16 s

TO FIND:

Distance covered by the object = 2080 m

EXPLANATION:

$\boxed{\bold{\large{S = ut + \dfrac{1}{2}at^2}}}$

u = 50 m/s

t = 16 s

a = 10 m/s²

S = 50(16) +1/2 10(16²)

S = 800 + 5 × 256

S = 800 + 1280

S = 2080 m

∴ The distance covered by the object will be equal to 2080 m.

Some more formulae:

$\boxed{\bold{\large{v = u + at}}}$

$\boxed{\bold{\large{v^2 - u^2 =2 as}}}$