Physics Middle School

. A car moves forward up a hill at 12 m/s with a uni-
form backward acceleration of 1.6 m/s2.
a. What is its displacement after 6.0 s?
b. What is its displacement after 9.0 s?

Answers

Answer 1

A) Displacement after 6.0 s 43.2 m uphill.

B) Displacement after 9.0 s 43.2 m uphill.

Explanation:

A car moving upwards in a hill is [tex]12 ms^{-1}[/tex].

Its uniform backward acceleration is [tex]-1.6ms^{-2}[/tex]. (since backward acceleration is a negative acceleration, it is mentioned in negative)

We need to find the displacement of the car after some time.

Using the equation of the motion formula, we know can identify the displacement.

D=[tex]vt+\frac{1}{2} at^2[/tex].

a) Displacement after 6.0 seconds,

D = [tex]12(6.0)+\frac{1}{2}(-1.6)(6.0)^2[/tex].

=[tex]72+\frac{1}{2} (36)(-1.6).[/tex]

=[tex]72+\frac{1}{2}(-57.6).[/tex]

=72-28.8.

D=43.2 m.

b) Displacement after 9.0 seconds,

D= [tex]12(9.0)+\frac{1}{2}(-1.6)(9.0)^2[/tex].

=[tex]108+\frac{1}{2} (81)(-1.6).[/tex]

=[tex]108+\frac{1}{2}(-129.6).[/tex]

= 108-64.8.

D=43.2 m.

Answer 2

The car's displacement after 6.0 s is 14.4 m and after 9.0 s is 43.2 m.

To find the displacement of the car after a given time, we can use the equation:
Displacement (d) = Initial velocity (v) * time (t) + (1/2) * acceleration (a) * time^2

a. After 6.0 s:
Initial velocity (v) = 12 m/s
Acceleration (a) = -1.6 m/s^2 (negative because it's a backward acceleration)
Substituting the values into the equation:
d = (12 m/s) * (6.0 s) + (1/2) * (-1.6 m/s^2) * (6.0 s)^2 = 72 m - 57.6 m = 14.4 m

b. After 9.0 s:
Using the same equation and substituting the new time, we can calculate the displacement:
d = (12 m/s) * (9.0 s) + (1/2) * (-1.6 m/s^2) * (9.0 s)^2 = 108 m - 64.8 m = 43.2 m

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Related Questions

If the mass of an object is 44 kilograms and its velocity is 10 meters per second east, how much Kinetic Energy does it have?

Answers

Answer: 2200J

Explanation:

M = 44kg

V = 10m/s

K.E =?

K.E = 1/2MV2 = 1/2 x 44 x (10)^2

K.E = 22 x 100

K.E = 2200J

a child pulls on a string that is attached to a car. if the child does 80.2 J of work while pulling the car 25.0 m, with what force is the child pulling?

Answers

Answer:

F = 3.20 N

Explanation:

Given:

Work done by child = 80.2 j

Distance that the car moves = 25.0 m

We need to find the force acting on the car.

Solution:

Using work done formula as.

[tex]W = F\times d[/tex]

Where:

W = Work done by any object.

F = Force (push or pull)

d = distance that the object moves.

Substitute [tex]W = 80.2\ J\ and\ d =25.0\ m[/tex] in work done formula.

[tex]80.2 = F\times 25[/tex]

[tex]F=\frac{80.2}{25}[/tex]

F = 3.20 N

Therefore, force acting on the car F = 3.20 N

Which of the following is a result of gravitational forces in the Solar System?
A.
the radiation given off by Jupiter
B.
Saturn is further away from the Sun than Earth
C.
the difference in surface temperature on each of the planets
D.
the orbit of moons around their planets in the Solar system

Answers

Answer:

D

Explanation:

the answer is d because gravitational force is what allows them to rotate

hope this was helpful

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