Question

Equation of position of a particle is x=5t^2 where x is in meter and t is in sec. find velocity and acceleration of partical at t=2sec​

Answer 1

Answer:

V = final velocity = 20 m/s

a = acceleration = 10 m/s²

Explanation:

Given :

• Equation of motion of the particle = (x=5t²)

To find :

• Velocity and the acceleration of the particle at t=2 seconds

Substituting the value of t=2 in the equation, we can find the distance travelled by the particle in 2 seconds

X = 5 (2)²

X = 5 (4)

X = 20 metres

The distance travelled by the particle is 20 metres

Initial velocity at t = 0 = 5(0)² = 0 (As distance is 0 at starting, initial velocity is 0 too)

Time taken = t = 2 seconds

Distance travelled = s = 20 metres

Using second equation of motion :

S=ut+½×at²

20=0×2+½×a×2²

20=2a

a = 20/2

a = 10 m/s²

Now using the Third equation of motion :

V²-u²=2as

V²-0²=2×10×20

V²=400

V=20 m/s

The final velocity of the particle is 20 m/s and the acceleration of the particle is 10 m/s²

Answer 2

Answer :

Given -

• Position of the particle = x = 5t²
• x is in meters and t is in seconds

To Find -

• Velocity at 2 seconds ?
• Acceleration at 2 Seconds ?

Solution -

x = 5t²

Substituting t = 2 seconds to find the position of the particle.

⇒ x = 5 × 2²

⇒ x = 5 × 4

⇒ x = 20 m

Now, We will.use equations of motion to Find the values of velocity and acceleration.

s = ut + ½ at²

• s = 20 m
• u = 0 m/s
• t = 2 seconds

⇒ 20 = 0*2 + ½ *a*2²

⇒ 20 = 0 + 2a

⇒ 20 = 2a

⇒ a = 20/2

⇒ a = 10 m/s²

Now, v² - u² = 2as

• u = 0 m/s
• a = 10 m/s²
• s = 20 m

⇒ v² - 0² = 2*10*20

⇒ v² = 400

⇒ v = √400

⇒ v = 20 m/s

Hence, the value of velocity at t = 2 seconds is 20 m/s and the value of acceleration at t = 2 seconds is 10 m/s².