A cyclist travels a distance of 4km from a to b and then moves a distance of 3km towards north find distance and displacement
Given: AB = 4km[0o],
BC = 3km[90o], (should be 90o between BC and AB),
d = 4 + 3 = 7km.
Disp. = sqrt(4^2+3^2) =
Total distance = 4 km + 3 km = 7 km
Displacement = the area of a triangle as the figure is making a right angle triangle so
1/2 . B. H
1/2. 3 km . 4 km
= 6 km
A cyclist travels a distance of 4km from a to b and then moves a distance of 3km towards north find distance and displacement
geez - impatient much? repost after 4 minutes, and don't even fix you error?
distance clearly equals 4+3
displacement is impossible to say, because we don't know the direction from a to b. It could be anything from 1 to 7
To find the distance traveled by the cyclist, you need to add up the distances traveled from point A to point B and from point B to the north.
Given:
Distance from A to B = 4 km
Distance from B to the north = 3 km
To find the distance:
Total distance = Distance from A to B + Distance from B to the north
= 4 km + 3 km
= 7 km
So, the distance traveled by the cyclist is 7 km.
Now, let's find the displacement. Displacement refers to the straight-line distance between the initial and final positions.
Since the cyclist first travels from point A to point B, and then moves towards the north, the displacement will be equal to the straight-line distance between point A and the final position of the cyclist.
The displacement can be calculated using the Pythagorean theorem, where the distance traveled from A to B is the horizontal component, and the distance traveled towards the north is the vertical component.
Given:
Distance from A to B = 4 km
Distance from B to the north = 3 km
To find the displacement:
Displacement = √(Horizontal component^2 + Vertical component^2)
= √(4 km^2 + 3 km^2)
= √(16 km^2 + 9 km^2)
= √(25 km^2)
= 5 km
So, the displacement of the cyclist is 5 km.