Two hockey pucks, labeled A and B, are initially at rest on asmooth ice surface and are separated by a distance of18.0 m. Simultaneously, each puck is given a quickpush and they begin to slide directly toward each other. Puck Amoves with a speed of v_A = 3.50 m/s, and puck B moves with a speed ofv_B = 3.90 m/s.

What is the distance that puck A covers prior to the collision?

8.505 m

Explanation:

Let V1 and V2 be velocities of puck A and B respectively

Since A and B move in the same direction, so the relative velocity will be V1+V2=3.5+3.9=7.4m/s

Or

Vr=7.4 m/s

Distance=S= 18 m

Time =t=?

S=Vr×t

==> t=S/Vr

==> t= 18/7.4=2.43 sec

At this time both will strike together

Distance by puck A

V1=3.5 m/s

Time=t= 2.43 sec

Distance covered=d=?

d=V1×t=3.5×2.43=8.505 m

So, puck A will cover 8.505 meters before collision

total distance travelled = 60 miles

total time taken =1.5 hours

average speed = total distance / total time

60 / 1.5

= 40

therefore, average speed = 40 m/h

happy to

pls mark as brainliest

what was the acceleration of the 2 kg cart?

⇒ 16.0 cm/s2

what was the acceleration of the 4 kg cart?

⇒ 8.1 cm/s2

what was the acceleration of the 6 kg cart?

⇒ 5.3 cm/s2