The driver of a car travelling along a straight road with a speed of 72km/hour observe a signboard which give a speed limit to be 54 km/hour. The sign board is 70m ahead when the driver applies the break calculate the acceleration of the car which will cause the car to pass the signboard at the stated speed limit?
To calculate the acceleration of the car, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v is the final velocity (54 km/hour)
u is the initial velocity (72 km/hour)
a is the acceleration (to be calculated)
s is the distance traveled (70m)
First, we need to convert the velocities from km/hour to m/s:
54 km/hour = 54 * (1000/3600) m/s = 15 m/s
72 km/hour = 72 * (1000/3600) m/s = 20 m/s
Now, we can substitute the values into the equation:
(15 m/s)^2 = (20 m/s)^2 + 2a * 70m
225 m^2/s^2 = 400 m^2/s^2 + 140a
Rearranging the equation, we have:
140a = 400 m^2/s^2 - 225 m^2/s^2
140a = 175 m^2/s^2
Dividing both sides by 140, we get:
a = 175 m^2/s^2 / 140
a ≈ 1.25 m/s^2
Therefore, the acceleration of the car required to pass the signboard at the stated speed limit of 54 km/hour is approximately 1.25 m/s^2.
vf^2=vi^2+2ad
change 72km/hr, 54km/hr to m/s
then solve for a (in m/s)