The motion of a particle along a straight line is described by equati...
Problem: The motion of a particle along a straight line is described by equation x=8+12t-t³ where x is in metre and t is in Sec. The retardation of the particle when its velocity becomes zero is?
Solution:
To find the retardation of the particle, we first need to find the velocity of the particle. We can find the velocity by taking the derivative of the position equation with respect to time.
x = 8 + 12t - t³
v = dx/dt = 12 - 3t²
Setting the velocity to zero, we get:
12 - 3t² = 0
Solving for t, we get:
t = ±2
Since we are interested in the point where the velocity becomes zero, we take the positive value of t, which is t = 2 sec.
To find the retardation, we need to find the acceleration of the particle at t = 2 sec. We can find the acceleration by taking the derivative of the velocity equation with respect to time.
v = 12 - 3t²
a = dv/dt = -6t
Plugging in t = 2 sec, we get:
a = -12 m/s²
Therefore, the retardation of the particle when its velocity becomes zero is -12 m/s². This means that the particle is decelerating at a rate of 12 m/s².
The motion of a particle along a straight line is described by equati...
Answer:
12 m/s²
Explanation:
x = 8 + 12t – t³ The final velocity of the particle will be zero, because it retarded.
V= 0 + 12 – 3t² = 0 ⇒ 3t² = 12 ⇒ t = 2 sec
Now the retardation a = dv/ dt = 0 – 6t
a [ t= 2] = – 12 m/s²
retardation = 12 m/s²
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