A body moving with a velocity of 3m/s is brought to rest by a constant force after travelling 15m, calculate the retardation.
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To calculate the retardation, we can use the equation of motion that relates velocity, initial velocity, acceleration, and distance:
\(v^2 = u^2 + 2as\)
where:
v = final velocity (0 m/s, since the body comes to rest)
u = initial velocity (3 m/s)
a = acceleration (retardation in this case)
s = distance traveled (15 m)
Plugging in the given values:
\(0 = (3)^2 + 2a(15)\)
Simplifying the equation:
\(0 = 9 + 30a\)
Rearranging the equation in terms of \(a\):
\(30a = -9\)
Dividing both sides by 30:
\(a = -\frac{9}{30}\)
Simplifying:
\(a = -0.3\ m/s^2\)
Therefore, the retardation of the body is -0.3 m/s². The negative sign indicates that the acceleration is opposite to the initial velocity, causing the body to decelerate.
v = Vi + a t
0 = 3 + a t
t = -3/a
d = Vi t + (1/2) a t^2
15 = 3 t + (1/2) a t^2
15 = 3(-3/a) + (1/2) a 9/a^2
15 = -9/a + 9/2a = -9/2a
30 a = -9
a = -9/30