A small car of mass 833 kg is parked behind a small truck of mass 1767 kg on a level road. The brakes of both the car and the truck are off so that they are free to roll with negligible friction. A 41 kg woman sitting on the tailgate of the truck shoves the car away by exerting a constant force on the car with her feet. The car accelerates at 1.5 m/s 2 . What is the acceleration of the truck

the acceleration of the truck a_t is 0.6911 m/s²

Explanation:

Given the data in the question;

There is no external force on the system; net force on the system is 0

Mass of the truck with the woman M = 1767 kg + 41 kg = 1808 kg

mass of the car m = 833 kg

car acceleration a_c = 1.5 m/s²

now let a_t be the acceleration of the truck in opposite direction

Action force on the car = Reaction force on the car

ma_c = Ma_t

a_t = ma_c / M

we substitute

a_t = (833 × 1.5) / 1808

a_t = 1249.5 / 1808

a_t = 0.6911 m/s²    

Therefore, the acceleration of the truck a_t is 0.6911 m/s²

the duration of a photographic flash is related to an rc time constant, which is 0.100 μs for a certain camera.

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000