Question

A car traveling with constant speed travels 150 km in 7200 s. What is the speed of the car?
Express the speed to two significant figures and include the appropriate units.

Answer 1

The speed of the car is 75 km/h

Further explanation

These are the formulas that we have to remember before solving the problem.

Speed is the rate of change of distance.

$\large { \boxed {v = \frac{d}{t}}}$

v = speed ( m/s )

d = distance ( m )

t = time ( s )

Acceleration is the rate of change of velocity.

$\large { \boxed {a = \frac{\Delta v}{t}}}$

a = acceleration ( m/s² )

Δv = change in speed ( m/s )

t = time ( s )

Let us now tackle the problem!

Given:

distance = d = 150 km

time taken = t = 7200 s = 2 hours

Unknown:

velocity = v = ?

Solution:

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

$v = \frac{150}{2}$

$\large {\boxed {v = 75~km/h} }$

The acceleration of the car is 0 m/s² because it travels with constant speed.

We could also plot the distance vs time graph as shown in the attachment.

Learn more

• Velocity of A Runner : https://brainly.com/question/3813437
• Kinetic Energy : https://brainly.com/question/692781
• Acceleration : https://brainly.com/question/2283922

Answer details

Grade: Middle School

Subject: Physics

Chapter: Kinematics

Keywords: indycar top speed of a fastest police car has ever gone

Answer 2

Answer:

The speed of the car to 2 significant figures is $\boxed{21\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ .

Further Explanation:

Given:

The distance covered by the car is $150\,{\text{km}}$.

The time taken by the car to cover the distance is $7200\,{\text{s}}$ .

Concept:

The speed of a car defined as the ratio of the distance and the time taken by the car in covering that distance.

Write the expression for the speed of the car as:

 $v = \dfrac{{{\text{distance}}}}{{{\text{time}}}}$    …… (1)                                        

Here, $v$ is the speed of the car.

The distance covered by the car in meters is.

 $\begin{aligned}d&= 150 \times {\text{1000}}\,{\text{m}}\\{\text{ 1}}&={\text{.50}} \times {\text{1}}{{\text{0}}^5}\,{\text{m}}\\\end{aligned}$

Substitute the values in the above expression.

 $\begin{aligned}v&= \frac{{1.50 \times {{10}^5}\,{\text{m}}}}{{7200\,{\text{s}}}}\\&= \frac{{1500}}{{72}}\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\&= 20.83\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\&\approx 21\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}$

Thus, the speed of the car to 2 significant figures is $\boxed{21\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ .

Learn More:

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Answer Details:

Grade: Middle School

Chapter: Speed and distance

Subject: Physics

Keywords: Speed, velocity, distance, time, distance of 150 km, in 7200 s, speed of the car, ratio of distance, rate of change of distance, 20.83 m/s, 21m/s.