A car starts from rest and moves with constant acceleration during the 5th second of its motion,it covers a distance of 36 meters. what is the acceleration
since s = 1/2 at^2, then during 4 <=t <=5 (the 5th second)
36 = 1/2 a 5^2 - 1/2 a 4^2 = 9/2 a
a = 8
it means car have 8 m/s^2 constant acceleration.
Then what do you think steve if we put this acceleration to calculate exact time period which will be greater than 4 seconds?
Thnx a lot steve
If we have to find the total distance also then how we'll do it???
A car start from raised and move with constant acceleration during the five second of it motion it cover at distance of 36 M calculation the acceleration of car and total distance covered by the car during this time
Oh, look at that speedy car! It covered 36 meters during its fifth second of motion. To find the acceleration, we need to know the initial velocity and time as well. Did the car tell you any other details while it was zooming past?
To find the acceleration of the car, we need to use the kinematic equation that relates distance, initial velocity, time, and acceleration. The equation is:
distance = initial velocity * time + (0.5 * acceleration * time^2)
In this case, the car starts from rest, so the initial velocity is 0. We are given the distance (36 meters) and the time (5 seconds).
Plugging in the known values into the equation, we have:
36 = 0 * 5 + (0.5 * acceleration * 5^2)
Simplifying further:
36 = 0 + (0.5 * 25 * acceleration)
36 = 12.5 * acceleration
Dividing both sides by 12.5:
acceleration = 36 / 12.5
acceleration = 2.88 m/s^2
Therefore, the acceleration of the car is 2.88 m/s^2.