All categories

3 years ago

On a highway, a car is driven 80. kilometers

Physics

2 answers:

nydimaria [60]3 years ago

7 0

The average speed of the car for the entire trip can be calculate by using:

$v=\frac{S}{t}$

where S is the total distance covered by the car, and t is the total time taken.

The total distance travelled by the car is:

$S=80 km+50 km+40 km=170 km$

while the total time taken is:

$t=1.00 h+0.50 h+0.50 h=2.00 h$

so, the average speed of the car is:

$v=\frac{S}{t}=\frac{170 km}{2.00 h}=85 km/h$

so, the correct answer is (3) 85 km/h.

Bas_tet [7]3 years ago

4 0

The average speed of the car for entire trip is $\boxed{85\text{ km/h}}$ .

Further Explanation:

Speed is the measure of a quantity of an object the tells how fast the object is moving in the other words we can define the speed that it is the distance covered by an body divided by the time taken to cover that distance. It is a quantity with only magnitude so it is a scalar quantity.

The average speed is defined as the sum of all the distance traveled by the body divided by the sum of time taken to travel that distance.

Given:

The distance travel by the car is $80\text{ km}$ for $1.00\text{ hours}$ .

The distance travel by the car is $50\text{ km}$ for $0.50\text{ hours}$ .

The distance travel by the car is $40\text{ km}$ for .

$0.50\text{ hours}$ .

Concept:

The expression for the average can be written as:

S = $\dfrac{{{\text{sum of distances }}}}{{{\text{sum of time}}}}$ …… (1)

The distance travel by the car is $80\text{ km}$ for $1.00\text{ hours}$ , distance travel by the car is $50\text{ km}$ for $0.50\text{ hours}$ and distance travel by the car is $40\text{ km}$ for $0.50\text{ hours}$ .

The sum of the distance is:

$\begin{aligned}D&=80{\text{ km}}+{\text{50 km}}+40{\text{ km}}\\&=170{\text{ km}}\end{aligned}$

The sum of time taken to travel the distance is:

$\begin{aligned}t&=1.00{\text{ h}} + 0.50{\text{ h}} + 0.50{\text{ h}}\\&=2.00{\text{ h}}\end{aligned}$

Substitute $170\text{ km}$ for total distance and $2.00\text{ h}$ for total time in equation (1).

$\begin{aligned}S&=\frac{{{\text{170 km}}}}{{2.00{\text{ h}}}}\\&=85{\text{ km/h}}\end{aligned}$

Therefore, the average speed of the car for entire trip is $\boxed{85\text{ km/h}}$ .

Learn more:

1. Find the net force. https://brainly.in/question/8668644

2. Find the velocity. https://brainly.in/question/4502003

3. Calculate average speed brainly.com/question/11597590

Answer Details:

Grade: Middle school

Subject: Physics

Chapter: Kinematics

Keywords:

Highway, car, driven, 80 km, 1.00 hour, 1.00 hr, 1.00 h, 50 km, 0.50 hour, 0.50 hr, 0.50 h, 50 km, average speed, entire, trip, 85 km/hr.

You might be interested in

If you whirl a tin can on the end of a string and the string suddenly breaks, in what direction will the can go?

Contact [7]

Answer:

fly off, tangent to its circular path.

Explanation:

4 0

3 years ago

What is the correct equation kinetic enery

Ivahew [28]

Kinetic energy is directly proportional to the mass of the object and to the square of its velocity: K.E. = 1/2 m v2.

4 0

3 years ago

A beaker has a mass of 125g. What is the mass of this beaker in decigrams

Anit [1.1K]

1250 decigrams
1 gram = 10 decigrams

4 0

3 years ago

A 95kg fullback (football player for those not into sports) moving south with a speed of 5.0 m/s has a perfectly inelastic colli

Lunna [17]

Answer:

a. $v=3.11mls$ , $29.4^{0}$

b. $K.E =-697.8J$

Explanation:

To calculate the values in the question, a deep understanding of perfect inelastic collision is important.

When two bodies undergo inelastic collision, two important parameters must be well understood i.e

Momentum: the momentum is always conserved in perfectly inelastic collision. i.e the total momentum after collision is the sum of the individual momentum before collision

Kinetic energy: Kinetic energy is not conserved due to dissipative force.

a.To calculate the velocity, we first find the total momentum before collision

Momentum of player 1 $p_{1} =mv=95kg*5m/s\\p_{1} =475kgm/s\\$

Momentum of player 2 $p_{2} =mv=90kg*3m/s\\p_{1} =270kgm/s\\$

Hence the total momentum $p_{12}=p_{1}+p_{2}\\$

Note, since the direction of movement before collision is due south and due north respectively we have to represent the velocity using the rectangular coordinate

Hence $p_{12}=(m_{1}+m_{2})v=p_{1}i+p_{2}j\\$

$(95+90)v=475i+270j\\$

$v=2.57i+1.45j\\$

solving for the resultant velocity, we have

$v=\sqrt{2.75^{2} +1.45^{2}}\\ v=3.11mls$

To calculate the direction of movement, we have

$\alpha =tan^{-1}=\frac{v_{j} }{v_{i}}\\ \alpha =tan^{-1}=\frac{1.45}{2.57}\\\alpha =29.4^{0}$

b. to calculate the decrease in total kinetic energy, before collision, the total kinetic was

$K.E_{initial} =\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}.\\K.E_{initial} =((1/2)*95*5^{2})+((1/2)*90*3^{2})\\K.E_{initial} =1187.5+405\\K.E_{initial} =1592.5J\\$