A 2150-kg truck is traveling along a staight, level road at a constant speed of 55.0 km/h when the driver removes his foot from the accelerator. After 21.0 s, the truck's speed is 33.0 km/h. What is the magnitude of the average net force acting on the truck during the 21.0 s interval?
2007-06-25 14:28:06 · 2 answers · asked by Tam 1 in Science & Mathematics ➔ Physics
To find the average acceleration, subtract the two velocities and divide by time. You will have to convert your velocities to m/s:
55 - 33 = 22 *1000/3600 = 6.1 m/s
6.1 m/s / 21 s = 0.29 m/s^2
Multiply mass times acceleration to get force:
F = 2150 * 0.29 = 626 N
Darn! 1 second too late!
2007-06-25 14:34:47 · answer #1 · answered by Jeff 3 · 1⤊ 0⤋
F = ma
v1 = 55 km/h = 15.2777 m/s
v2 = 33 km/h = 9.1666 m/s
a = (9.16666 - 15.27777) / 21
a = -0.291 m/s^2
F = (2150)(-0.291)
F = -626 N
magnitude is 626 N
2007-06-25 14:34:46 · answer #2 · answered by yeeeehaw 5 · 0⤊ 0⤋