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Liono4ka [1.6K]
3 years ago
12

I. What is the initial velocity of the car?

Physics
1 answer:
qaws [65]3 years ago
6 0

Answer:

I. 0 m/s

II. 20 m/s

III. Part BC

Explanation:

I. Determination of the initial velocity.

From the diagram given above,

The motion of the car starts from the origin. This implies that the car start from rest and as such, the initial velocity of the car is 0 m/s

II. Determination of the maximum velocity attained.

From the diagram given above, we can see clearly that the maximum velocity is 20 m/s.

III. Determination of the part of the graph that represents zero acceleration.

It important that we know the meaning of zero acceleration.

Zero acceleration simply means the car is not accelerating. This can only be true when the car is moving with a constant velocity.

From the graph given above, the car has a constant velocity between B and C.

Therefore, part BC illustrates zero acceleration.

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Please need help fast
iVinArrow [24]

(a) See graph in attachment

The appropriate graph to draw in this part is a graph of velocity vs time.

In this problem, we have a horse that accelerates from 0 m/s to 15 m/s in 10 s.

Assuming the acceleration of the horse is uniform, it means that the velocity (y-coordinate of the graph) must increase linearly with the time: therefore, the velocity-time graph will appear as a straight line, having the final point at

t = 10 s

v = 15 m/s

(b) 1.5 m/s^2

The average acceleration of the horse can be calculated as:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time interval

In this problem,

u = 0

v = 15 m/s

t = 10 s

Substituting,

a=\frac{15-0}{10}=1.5 m/s^2

(c) 75 m

For a uniformly accelerated motion, the distance travelled can be calculated by using the suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance travelled

u is the initial velocity

t is the time interval

a is the acceleration

In this problem,

u = 0

t = 10 s

a=1.5 m/s^2

Substituting,

s=0+\frac{1}{2}(1.5)(10)^2=75 m

(d) See attached graphs

In a uniformly accelerated motion:

- The distance travelled (x) follows the equation mentioned in part c,

x=ut+\frac{1}{2}at^2

So, we see that this has the form of a parabola: therefore, the graph x vs t will represents a parabola.

- The acceleration is constant during the motion, and its value is

a=1.5 m/s^2 (calculated in part b)

therefore, the graph acceleration vs time will show a flat line at a constant value of 1.5 m/s^2.

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