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3 years ago

A car drives 30 m north in 1.5 seconds and drives 60 east in 2 seconds. What is the average velocity

Physics

1 answer:

Bad White [126]3 years ago

4 0

Answer:

19.2 m/s

Explanation:

From the question,

Velocity = displacement/time = d/t

V = d/t.................................. Equation 1

Average velocity = Total displacement/ total time

Total dispalcement = √(30²+60²) = √4500 = 67.08 m

Total time = 1.5+2 = 3.5 s.

Substitute these value into equation 1

V = 67.08/3.5

V = 19.2 m/s.

Hence the average velocity is 19.2 m/s

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Explanation:

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3 years ago

A school bus takes 0.530 h to reach the school from your house. If the average velocity of the bus is 19.0 km/h to the east, wha

kifflom [539]

Answer:

$x_{distance}=10.07km$

Explanation:

Given data

time=0.530 h

Average velocity Vavg=19.0 km/s

To find

Displacement Δx

Solution

The Formula for average velocity is given as

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4 years ago

Help me find the acceleration

ANEK [815]

a = 3.09 m/s²

<h3>Explanation</h3>

This question doesn't tell anything about how long it took for the car to go through 105 meters. As a result, the timeless suvat equation is likely what you need for this question.

In the timeless suvat equation,

$a = \dfrac{v^2 - u^2}{2\; x}$

where

$a$ is the acceleration of the car;
$v$ is the final velocity of the car;
$u$ is the initial velocity of the car; and
$x$ is the displacement of the car.

Note that v and u are velocities. Make sure that you include their signs in the calculation.

In this question,

$a$ is the unknown;
$v = -10.9 \; \text{m} \cdot \text{s}^{-2}$ ;
$u = -27.7 \; \text{m} \cdot \text{s}^{-2}$ ; and
$x = - 105 \; \text{m}$ .

Apply the timeless suvat equation:

$a = \dfrac{v^{2} - u^{2}}{2\; x}\\\phantom{a} = \dfrac{(-10.9)^{2} - (-27.7)^{2}}{2 \times (-105)}\\\phantom{a} = 3.09 \; \text{m} \cdot \text{s}^{-2}$ .

The value of $a$ is greater than zero, which is reasonable. Velocity of the car is negative, meaning that the car is moving backward. The car now moves to the back at a slower speed. Effectively it accelerates to the front. Its acceleration shall thus be positive.

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