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VMariaS [17]
3 years ago
9

A car accelerates from rest at -3.00m/s^2. What is the velocity at the end of 5.0s? What is the displacement after5.0s?

Physics
1 answer:
Andrew [12]3 years ago
5 0

Speed = (acceleration) x (time)
Velocity = (speed) in (direction of the speed)

Speed = (-3 m/s²) x (5 s) = 15 m/s
Velocity =
             (15 m/s) in the direction opposite to the direction you call positive
.

Displacement = (distance between start-point and end-point)
                           in the direction from start-point to end-point.

Distance = (1/2) (acceleration) (time)²
Distance = (1/2) (3 m/s²) (5 s)²
                 = (1/2) (3 m/s²) (25 s²)  =  37.5 meters

Displacement =
                     37.5 meters in the direction opposite to the direction you call positive.

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A rod of propoer length l0 is at rest in frame S'. it ilies in the x',y' plane and amkes an angle of sin^-1(3/5) What must be th
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Answer:

v = 1.98*10^8 m/s

Explanation:

Given:

- Rod at rest in S' frame

- makes an angle Q = sin^-1 (3/5) in reference frame S'

- makes an angle of 45 degree in frame S

Find:

What must be the value of v if as measured in S the rod is at a 45 degree)

Solution:

- In reference frame S'

                x' component = L*cos(Q)

                y' component = L*sin(Q)

- Apply length contraction to convert projected S' frame lengths to S frame:

                x component = L*cos(Q) / γ           (Length contraction)

                y component = L*sin(Q)                  (No motion)

- If the rod is at angle 45° to the x axis, as measured in F, then the x and y components must be  equal:

                L*sin(Q) = L*cos(Q) / γ    

Given:       γ = c / sqrt(c^2 - v^2)

                 c / sqrt(c^2 - v^2) = cot(Q)

                 1 - (v/c)^2 = tan(Q)

                 v = c*sqrt( 1 - tan^2 (Q))

For the case when Q = sin^-1 (3/5)::

                 tan(Q) = 3/4

                 v = c*sqrt( 1 - (3/4)^2)

                 v = c*sqrt(7) / 4 = 1.98*10^8 m/s

 

5 0
3 years ago
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