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pashok25 [27]
3 years ago
10

A car moves with an initial velocity of 19 m/s due north. (Part A) Find the velocity of the car after 5.6 s if its acceleration

is 1.6 m/s^2 due north. (Part B) Find the velocity of the car after 5.6 s if its acceleration is 1.5 m/s2 due south.
Physics
1 answer:
galben [10]3 years ago
5 0

Answer

Assuming

east is the positive x direction

north is the positive y direction

initial velocity , u = 19 j m/s

a) acceleration , a = 1.6 j m/s^2

Using first equation of motion

v = u + a × t

v = 19 + 5.6× 1.6

v = 28 j m/s

the velocity of the car after 5.6 s is 28 m/s north

b)

acceleration , a = -1.5 j m/s^2

Using first equation of motion

v = u + a × t

v = 19 - 5.6 ×1.5

v = 10.6 j m/s

the velocity of the car after 5.6 s is 10.6 m/s north

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An object is originally moving at a constant velocity of 8 m/s in the -x direction. It moves at this constant velocity for 3 sec
aivan3 [116]

Answer:

244.64m

Explanation:

First, we find the distance traveled with constant velocity. It's simply multiplying velocity time the time that elapsed:

x = V*t = -8\frac{m}{s} *3s = -24m

After this, the ball will start traveling with a constant acceleration motion. Due to the fact that the acceleration is the opposite direction to the initial velocity, this motion will have 2 phases:

1. The velocity will start to decrease untill it reaches 0m/s.

2. Then, the velocity will start to increase at the rate of the acceleration.

The distance that the ball travels in the first phase can be found with the following expression:

v^2 = v_0^2 + 2a*d

Where v is the final velocity (0m/s), v_0 is the initial velocity (-8m/s) and a is the acceleration (+9m/s^2). We solve for d:

d = \frac{v^2 - v_0^2}{2a} = \frac{(0m/s)^2 - (-8m/s)^2}{2*7m/s^2}= -4.57m

Now, before finding the distance traveled in the second phase, we need to find the time that took for the velocity to reach 0:

t_1 = \frac{v}{a} = \frac{8m/s}{7m/s^2} = 1.143 s

Then, the time of the second phase will be:

t_2 = 9s - t_1 = 9s - 1.143s = 7.857s

Using this, we using the equations for constant acceleration motion in order to calculate the distance traveled in the second phase:

x = \frac{1}{2}a*t^2 + v_0*t + x_0

V_0, the initial velocity of the second phase, will be 0 as previously mentioned. X_0, the initial position, will be 0, for simplicity:

x = \frac{1}{2}*7\frac{m}{s^2}*t^2 + 0m/s*t + 0m = 216.07m

So, the total distance covered by this object in meters will be the sum of all the distances we found:

x_total = 24m + 4.57m + 216.07m = 244.64m

8 0
3 years ago
A rocket travels 1.3 km in 62 ms. What is its average speed in m⋅s−1? Do not give your answer in scientific notation. The answer
hjlf

Answer:

Average speed = 0.35 m/s

Explanation:

Given the following data;

Distance = 1.3 Km

Time = 62 minutes

To find the average speed in m/s;

First of all, we would convert the quantities to their standard unit (S.I) of measurement;

Conversion:

1.3 kilometres to meters = 1.3 * 1000 = 1300 meters

For time;

1 minute = 60 seconds

62 minutes = X

Cross-multiplying, we have;

X = 62 * 60

X = 3720 seconds

Now, we can calculate the average speed in m/s using the formula;

Speed = \frac {distance}{time}

Speed = \frac {1300}{3720}

Average speed = 0.35 m/s

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