Derive first equation of motion by graphical method. The brakes applied to a car produce an acceleration of 6 m s-2in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time.
Answers
Answer 1st
~ Explanation:
Considering a open object that moves under uniform acceleration where u is not equal to zero from the graph we are able to see that the initial velocity of the object that is at point A and then it increased to final values that is at point B in the time, the velocity change at the uniform rate acceleration. In the graph the perpendicular line BC and BE are drawn from point B on the time and validity axis respectively.
~ Now according to the graph,
⇢ AO = DC = u (Initial velocity)
⇢ AD = OC = t (Time)
⇢ EO = BC = v (Final velocity)
~ Now let us derive velocity-time relation first. Let's do it.
Firstly we can write BC as BD + DC. Now as BD have the velocity position and DC have the time position. Henceforth, we already know that
As we write BD at the place of v-u henceforth,
Answer 2nd
Provided that:
- Acceleration = -6 m/s sq.
- Final velocity = 0 m/s
- Time taken = 2 seconds
- Distance = ???
Don't be confused! Acceleration cames in negative because it accelerated in opposite direction of the motion
Solution:
- Distance = 12 metres
Required solution:
~ Firstly finding initial velocity!
→ v = u + at
→ 0 = u + (-6)(2)
→ 0 = u + (-12)
→ 0 = u - 12
→ 0 + 12 = u
→ 12 = u
→ u = 12 m/s
→ Initial velocity = 12 m/s
~ Now let us find distance!
→ 2as = v² - u²
→ 2(-6)(s) = (0)² - (12)²
→ 2(-6)(s) = 0 - 144
→ 2(-6)(s) = -144
→ -12s = -144
→ 12s = 144
→ s = 144/12
→ s = 12 m
→ Distance = 12 metres
