From a speed of 75 kph, a car decelerates at the rate of 500 m/min^2 along a straight line.How far in meters, will it travel in 45 sec?
Firstly, we need to convert the deceleration rate from m/min^2 to m/s^2.
1 m/min^2 is equal to 0.01667 m/s^2.
Therefore, the deceleration rate is -8.33 m/s^2 (negative sign indicates deceleration).
Next, we convert the speed from kph to m/s.
1 kph is equal to 0.27778 m/s.
So, the initial speed of 75 kph is equal to 20.833 m/s.
Now, we can use the equation of motion to find the distance traveled:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, as the car comes to a stop)
u = initial velocity (20.833 m/s)
a = acceleration/deceleration (-8.33 m/s^2)
s = distance traveled
Substitute the values into the equation:
0 = (20.833)^2 + 2*(-8.33)*s
0 = 434.93 - 16.66*s
16.66*s = 434.93
s = 26.09 m
Therefore, the car will travel 26.09 meters in 45 seconds.