A car moves uniformly at a rate of 10 m/s square from an initial velocity of 18 km /h for 30 sec. Find the distance covered during this period ?
Answer the question
u = 18,000 m / 3600 s + 10 t = 5 +10 t
x = u initial t + (1/2) a t^2 = 5 (30) + (1/2)(10)(30)^2
= 150 + 5 * 900
I am expecting a raise by the way.
To find the distance covered during the 30 second period, we need to use the formula:
Distance = Initial velocity * Time + (1/2) * Acceleration * Time^2
First, we need to convert the initial velocity from km/h to m/s. To do this, divide the value by 3.6 (1 km/h = 0.2778 m/s):
Initial velocity = 18 km/h = 18 * 0.2778 m/s = 5 m/s
The acceleration is given as 10 m/s^2. Now we can substitute the values into the formula to find the distance covered:
Distance = 5 m/s * 30 s + (1/2) * 10 m/s^2 * (30 s)^2
= 150 m + (1/2) * 10 m/s^2 * 900 s^2
= 150 m + 5 m/s^2 * 900 s^2
= 150 m + 4500 m
= 4650 m
Therefore, the distance covered by the car during the 30 second period is 4650 meters.