Sally walks 10 meters forward and then 5 meters backwards. a) what is sally’s displacement? give magnitude (#) and direction – displacement is a vector. b) what is sally’s distance? give magnitude (#) only – distance is a scalar.

consider the forward direction as positive and backward direction as negative. hence displacement in froward direction will be taken as positive and displacement in backward direction will be taken as negative.

= displacement in forward direction = 10 meters

= displacement in backward direction = - 5  meters

= net displacement

net displacement is given as

= +

= 10 + (- 5)

=  5 meters

Since net displacement is positive , and forward is positive direction of motion.

hence the displacement is 5 meters in forward direction.

b)

d₁ = distance traveled in forward direction = 10 meters

d₂ = distance traveled in backward direction = 5 meters

D = total distance traveled

total distance traveled is given as

D = d₁ + d₂

D = 10 + 5

D = 15 meters

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.