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
Answers
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 :
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.