Question

The table shows test data for a corvette, one of the fastest cars made in the US. The car is moving in a straight line (the x-axis).. . Time (s) 0/2.10/20.0/53.0. Speed (mi/h)0/60.0/210/255. 1. Calculate the car's average acceleration (in m/s2) between 2.1 s and 20.0 s.. 2. Calculate the car's average acceleration (in m/s2) between 20.0 s and 53 s.

Answer 1

The car’s average acceleration between $2.1 s$ and $20.0 s$ is $\fbox{\begin\\{3.7468 m/\text{s}^2}\end{minispace}}$.

The car’s average acceleration between $20.0 s$ and $53.0 s$ is $\boxed{0.6095\text{ m}/\text{s}^2}$.

Further Explanation:

The acceleration is the rate of change in the velocity that is difference between the final velocity and initial velocity divided by the difference in the final time and initial time.

If the initial velocity of the object is less than the final velocity then the body must accelerate but when the initial velocity of the object is more than the final velocity then the body must decelerate.

Given:

The data for fastest cars made in U.S. in terms of time (s) and speed (mi/h) is:

At $0 s$ speed is $0 mi/h$, $2.10 s$ speed  is $60.0 mi/h$, $20.0 s$ speed is $210 mi/h$, $53.0 s$ speed is $255 mi/h$.

Concept:

The speed in mi/h can be written in terms of m/s as:

$1 mil/s=0.447 m/s$

The expression for the average acceleration is:

$a = \dfrac{{v - u}}{{T - t}}$

Here, $a$ is the average acceleration, $v$ is the final velocity, $u$  is the initial velocity, $t$  is the initial time and $T$ is the final time.

Part (1)

Substitute $60.0 mi/h$ for $u$, $210 mi/h$ for $v$, $2.10 s$ for $T$ and $20.0 s$ for $t$ in the above equation.

$\begin{aligned}a&=\frac{{\left( {210 - 60} \right)\left( {0.447{\text{ m/s}}} \right)}}{{\left( {20 - 2.1} \right){\text{ s}}}}\\&=\frac{{150\left( {0.447{\text{ m/s}}} \right)}}{{17.9{\text{ s}}}}\\&=3.7468{\text{ m/s}}\\\end{aligned}$

Therefore, the car’s average acceleration between $2.1 s$ and $20.0 s$ is $\boxed{3.7468 {{ m}/\text{s}^2}}$.

Part (2)

Substitute $210 mi/h$ for $u$, $255 mi/h$ for $v$, $20 s$ for $T$ and $53.0 s$ for $t$ in the above equation.

$\begin{aligned}a&=\frac{{\left( {255 - 210} \right)\left( {0.447{\text{ m/s}}} \right)}}{{\left( {53 - 20} \right){\text{ s}}}}\\&=\frac{{45\left( {0.447{\text{ m/s}}} \right)}}{{33{\text{ s}}}} \\&=0.6095{\text{ m/s }}\end{aligned}$

Therefore, the car’s average acceleration between $20.0 s$ and $53.0 s$ is $\boxed{0.6095 { m}/\text{s}^2}$.

Learn more:

1.  Average consumption of fuel brainly.com/question/3959122.

2.  Example of energy brainly.com/question/1062501.

3. Angular speed brainly.com/question/9575487.

Answer Details:

Grade: High school

Subject: Physics

Chapter: Kinematics

Keywords:

Table, corvette, fastest car, U.S., straight line, x-axis, time (s) 0 s, 2.10 s, 20.0 s, 53.0 s, speed (mil/h), 0 mil/h, 60 mil/h, 210 mil/h, 255 mil/h, 3.7468 m/s, 0.6095 m/s.

Answer 2

Answer

part 1) 3.746 m/s²

Part 2) 0.6096 m/s²

Explanation

Acceleration is the rate of change of speed.

It is calculated as follows:

Acceleration = Change in speed/change in time

a = (v - u)/t

Part 1

1mi/h = 0.44704 m/s

60 mi/h = 60 × 0.44704

= 26.8224 m/s

210 mi/h = 210 × 0.44704

= 93.8784 m/s

a = (93.8784 -26.8224 )/(20 - 2.1)

= 67.056/17.9

= 3.746 m/s

Part 2

1mi/h = 0.44704 m/s

255 mi/h = 255 × 0.44704

= 113.995 m/s

a = (113.995 - 93.8784)/(53 - 20)

=20.1166/33

= 0.6096 m/s²