The speed of the car is 75 km/h
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.
Grade: Middle School
Subject: Physics
Chapter: Kinematics
Keywords: indycar top speed of a fastest police car has ever gone
Answer:
The speed of the car to 2 significant figures is \(\boxed{21\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-
ulldelimiterspace} {\text{s}}}}\) .
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.
\(\begin{aligned}v&= \frac{{1.50 \times {{10}^5}\,{\text{m}}}}{{7200\,{\text{s}}}}\\&= \frac{{1500}}{{72}}\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right.\kern- ulldelimiterspace} {\text{s}}}\\&= 20.83\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern- ulldelimiterspace} {\text{s}}}\\&\approx 21\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern- ulldelimiterspace} {\text{s}}}\\\end{aligned}\)
Thus, the speed of the car to 2 significant figures is \(\boxed{21\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-
ulldelimiterspace} {\text{s}}}}\) .
Learn More:
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.
