A car takes off from rest and covers a distance of 8m on a straight road in log.calculate the magnitude of its acceleration
To calculate the magnitude of the car's acceleration, we need to determine its final velocity and the time it takes to reach that velocity.
Assuming the car starts from rest, its initial velocity (u) is 0 m/s.
Using the equation v^2 = u^2 + 2as, where v is the final velocity (unknown), s is the distance covered (8m), and a is the acceleration (unknown), we can rearrange the equation to solve for a.
v^2 = u^2 + 2as
v^2 = 0 + 2a(8m)
v^2 = 16a
Since the car is traveling in a straight line, there is no change in direction, so we can ignore the negative sign.
Now, we need to determine the time it takes for the car to travel the distance of 8m. We can use the equation v = u + at, where v is the final velocity (unknown), u is the initial velocity (0m/s), a is the acceleration (unknown), and t is the time.
v = u + at
v = 0 + at
v = at
Since the car has traveled a distance of 8m, we can substitute this value in place of s:
8m = (1/2)at^2
16 = at^2
Now, we have two equations:
v^2 = 16a
16 = at^2
Solving these equations simultaneously will give us the values of a and t:
16 = (v/t)^2
16 = a*t^2
Dividing the second equation by the first:
(v^2 / t^2) = a
Plugging in the value of v^2 from the first equation:
(16/t^2) = a
Simplifying:
a = 16/t^2
Since the value of t is not given in the problem, we cannot find the exact magnitude of the acceleration.