A motor car is uniformly retarded and brought to rest from a velocity 36km/hr in 5 seconds. Find its retardation and the distance covered during this period
To find the retardation of the car, we can use the following equation of motion:
\(v = u + at\)
where:
- \(v\) is the final velocity (0 m/s, as the car is brought to rest),
- \(u\) is the initial velocity (36 km/hr), and
- \(t\) is the time taken (5 seconds).
First, let's convert the initial velocity from km/hr to m/s. We know that:
1 km/hr = 1000 m/3600 s = 5/18 m/s
So, the initial velocity, \(u\), becomes:
\(u = 36 \times \frac{5}{18} \, \text{m/s}\)
Next, rearranging the equation to solve for acceleration, \(a\), we have:
\(a = \frac{v - u}{t}\)
Substituting the values we have:
\(a = \frac{0 - 36 \times \frac{5}{18}}{5} \, \text{m/s}^2\)
Simplifying the equation, we find:
\(a = -\frac{36 \times 5}{18 \times 5} \, \text{m/s}^2\)
Therefore, the retardation of the car is \(-\frac{10}{3} \, \text{m/s}^2\).
To calculate the distance covered during this period, we need to use the equation of motion:
\(s = ut + \frac{1}{2}at^2\)
Plugging in the given values:
\(s = (36 \times \frac{5}{18}) \times 5 + \frac{1}{2} \times (-\frac{10}{3}) \times (5)^2 \, \text{m}\)
Simplifying the equation, we get:
\(s = 25 \, \text{m}\)
Therefore, the distance covered by the car during this period is 25 meters.