Chapter 2, Problem 52P

(a)

To determine

The average acceleration of the car in second and in fourth gear.

Answer to Problem 52P

Acceleration in the second gear is $2{\text{m/s}}^{\text{2}}$ . Acceleration in the second gear is $1{\text{m/s}}^{\text{2}}$ .

Explanation of Solution

Given:

The given graph is shown below.

  

After applying second gear, the car starts driving in second gear at co-ordinates $\left(3,13\right)$ and ends at $\left(8,23\right)$ .

After applying fourth gear, car starts driving in fourth gear at co-ordinates $\left(18,35\right)$ and end at $\left(28,45\right)$ .

Formula used:

The acceleration is given by the formula,

  $a=\frac{v_2-v_1}{t_2-t_1}$

Calculation:

Acceleration after applying second gear is:

  $\begin{array}{c}a=\frac{23-13}{8-3}\\ =2{\text{m/s}}^{\text{2}}\end{array}$

Acceleration after applying fourth gear is:

  $\begin{array}{c}a=\frac{45-35}{28-18}\\ =1{\text{m/s}}^{\text{2}}\end{array}$

Conclusion:

Acceleration in the second gear is $2{\text{m/s}}^{\text{2}}$ .

Acceleration in the fourth gear is $1{\text{m/s}}^{\text{2}}$ .

(b)

To determine

The distance travelled by the car in fourth gear.

Answer to Problem 52P

Distance travelled by the car in fourth gear is 400 m.

Explanation of Solution

Given:

Velocity at the starting of fourth gear is $35\text{m/s}$ . Velocity at the end of fourth gear is $45\text{m/s}$ . Acceleration in the fourth gear is $1{\text{m/s}}^{\text{2}}$ .

The given graph is shown below.

  

Formula Used:

Using the third equation of motion $v^2-u^2=2as$

Calculation:

Distance travelled by car in fourth gear is:

  $\begin{array}{c}s=\frac{{\left(45\right)}^2-{\left(35\right)}^2}{2\times 1}\\ =\frac{2025-1225}{2}\\ =400\text{m}\end{array}$

Conclusion:

Distance travelled by car in fourth gear is 400 m.