A truck travels to and from a stone quarry that is located 2.5 km to the east. What is the total distance traveled by the truck? .00025 What is its displacement? .00025

5 km

0

Explanation:

The question says the truck travelled to and fro from the stone quarry.

When going to the stone quarry, the distance was 2.5 km to the east.

On returning, the distance back is also 2.5km

Therefore total distance travelled is 2.5km + 2.5km

= 5km

The displacement 8s dependent on what is the initial and also the final position. The initial and the final position are one and the same thing. Because the truck starts and goes back to the same position, the displacement is zero.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft