A motor car is uniformly retarded and brought to rest from a velocity 36km/h in 5s find it's retardation and the distance covered during the period
To find the retardation of the car, we can use the formula:
\(a = \frac{v-u}{t}\),
where:
\(a\) represents the retardation (acceleration in the opposite direction),
\(v\) represents the final velocity (0 m/s, since the car is brought to rest),
\(u\) represents the initial velocity (36 km/h),
\(t\) represents the time taken to come to rest (5 seconds).
First, we need to convert the initial velocity from km/h to m/s:
\(u = \frac{36 \, \text{km/h}}{3.6}\),
\(u = 10 \, \text{m/s}\).
Now, we can substitute the known values into the formula to find the retardation:
\(a = \frac{v - u}{t} = \frac{0 - 10}{5}\),
\(a = -2 \, \text{m/s}^2\) (negative sign indicates retardation).
So, the retardation of the car is \(2 \, \text{m/s}^2\) in the opposite direction.
To calculate the distance covered during the period, we need to use the equation:
\(s = ut + \frac{1}{2}at^2\),
where:
\(s\) represents the distance covered,
\(u\) represents the initial velocity,
\(t\) represents the time taken,
\(a\) represents the retardation.
Substituting the given values into the equation:
\(s = (10 \, \text{m/s})(5 \, \text{s}) + \frac{1}{2}(-2 \, \text{m/s}^2)(5 \, \text{s})^2\),
\(s = 50 \, \text{m} - 25 \, \text{m}\),
\(s = 25 \, \text{m}\).
Hence, the car covers a distance of \(25 \, \text{m}\) during the period.