Answer: a = 1.32m/s2
Therefore, the average acceleration is 1.32m/s2
Explanation:
Acceleration is the rate of change in the velocity per time
a = change in velocity/time
a = ∆v/t
average acceleration a = (v2 -v1)/t. ....1
Given;
Final velocity v2 = 1.63m/s
Initial velocity v1 = -1.15ms
time taken t = 2.11s
Substituting into eqn 1
a = [1.63 - (-1.15)]/2.11
a = (1.63+1.15)/2.11
a = 2.78/2.11
a = 1.32m/s2
Therefore, the average acceleration is 1.32m/s2
Assume a maximum stopping acceleration of g/2 where g is acceleration due to gravity.
Answer:
2.99 m/s
Explanation:
Stopping distance, s = 3 ft = 0.914 m
final velocity, v = 0
a = g/2 = 4.9 m/s²
Use third equation of motion:
substitute the values to find the speed of train:
The candle flame releases hot gases, which directly go in upwards directions. Due to which the air near the flame of the candle is very hot and dense. The particles along with vapour move up. And since the sideways, the air is not very dense and hot, we are able to hold the candle. In anti-gravity region, there will be no density differences and also, the convection process wont occur. So, the candle quickly snuffs off.
