Question

A PARTICLE IS MOVING TOWARDS EAST WITH VELOCITY 8M/S AND ACC IS 4M/S^2 TOWARDS WEST.FIND THE DISTANCE TRAVELLED IN 6 SECONDS

Answer 1

Answer:

Distance=40m

Displacement=-24m

Explanation:

Time when oblject will be at rest,

u=8m/s

a=-4m/s^2

v=0( body will come at rest)

Applying equations of motion

V=U+AT

0=8-4t

Therefore, t=2s

The body will come at rest after 2 seconds

Now, the distance travelled by the body to come at rest,

Applying 2nd equations of motion

S=UT+1/2AT^2

s=8×2-1/2×4×4

s=16-8=8m

Now, the distance travelled by the object in remaining 4 seconds

u=0m/s

a=-4m/s^2

t=4s

Applying 2nd equations of motion

s=0×4-1/2×4×16

s=-32m

Therefore total distance travelled is ,8+32 (distance can never be negative)=40m

Displacement=8-32=-24m