Question

A particle accelerates from rest at a constant rate for some time and attains a velocity of 8m/s. Afterwards it deaccelerates with the constant rate and comes to rest .If the total time taken is 4sec, what is the distance travelled?​

Answer 1

GIVEN :

• Initial velocity, u = 0 m/s.
• Final velocity, v = 8 m/s.
• time, t = 4 sec.

TO FIND :

• Distance travelled = ?

CONCEPTS USED :

• Newton's first law of motion.
• Newton's third law of motion.

SOLUTION :

First we have to find the acceleration (a) = ?

Using Newton's first law of motion,

V = u + at

=> 8 = 0 + a × 4

=> 8 = 4a

=> a = 8/4

$\qquad \large {\boxed {\sf a \ = \ 2 \ m/s^{2}}}$

Now,

To find the distance travelled (S) = ?

Using the Newton's third law of motion,

V² = u² + 2as

=> 8² = 0² + 2 × 2 × S

=> 64 = 0 + 4 × S

=> 64 = 4 S

=> S = 64/4

$\qquad \large {\boxed {\sf Distance \ = \ 16 \ m}}$

•°• The distance travelled is 16 meters.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

★ ALSO KEEP IN MIND :

• Newton's first law of motion : V = u +at.
• Newton's second law of motion : F = ma.
• Newton's third law of motion : V² = u² + 2as.

Answer 2

GIVEN :-

• Initial velocity, u = 0 m/s.
• Final velocity, v = 8 m/s.
• time, t = 4 sec.

TO FIND :-

• Distance travelled = ?

CONCEPTS USED :-

• Newton's first law of motion.
• Newton's third law of motion.

SOLUTION :-

First we have to find the acceleration (a) = ?

Using Newton's first law of motion,

V = u + at

=> 8 = 0 + a × 4

=> 8 = 4a

=> a = 8/4

$\qquad \large {\boxed {\sf a \ = \ 2 \ m/s^{2}}}$

Now,

To find the distance travelled (S) = ?

Using the Newton's third law of motion,

V² = u² + 2as

=> 8² = 0² + 2 × 2 × S

=> 64 = 0 + 4 × S

=> 64 = 4 S

=> S = 64/4

$\qquad \large {\boxed {\sf Distance \ = \ 16 \ m}}$

•°• The distance travelled is 16 meters.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

★ ALSO KEEP IN MIND :-

Newton's first law of motion : V = u +at.

Newton's second law of motion : F = ma.

Newton's third law of motion : V² = u² + 2as.