Chapter 3, Problem 57A

To determine

To Describe:Magnitude of acceleration from least to greatest.

Answer to Problem 57A

  $C<E<A<D<B$

Explanation of Solution

Given:

A.

Initial velocity, $u_A=0.50\text{m/s}$

Final velocity, $v_A=10.3\text{m/s}$

Time elapsed, $t_A=1.0\text{s}$

B.

Initial velocity, $u_B=0.40\text{cm/s}$

Final velocity, $v_B=0$

Time elapsed, $t_B=1.0\text{s}$

C.

Initial velocity, $u_C=0.40\text{cm/s}$

Final velocity, $v_C=2.0\text{cm/s}$

Time elapsed, $t_C=0.50\text{s}$

D.

Initial velocity, $u_D=0$

Final velocity, $v_D=4.3\text{m/s}$

Time elapsed, $t_D=0.40\text{s}$

E.

Initial velocity, $u_E=2.0\text{m/s}$

Final velocity, $v_E=1.0\text{m/s}$

Time elapsed, $t_E=8.3\text{s}$

Formula Used:

From first equation of motion:

  $\begin{array}{l}v=u+at\\ a=\frac{v-u}{t}\end{array}$

Where, v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.

Calculations:

Substitute the values and solve for acceleration:

A.

  $\begin{array}{c}{a}_A=\frac{v_A-u_A}{t_A}\\ =\frac{10.3-0.50}{1.0}\\ =9.8{\text{m/s}}^{\text{2}}\end{array}$

B.

  $\begin{array}{c}{a}_B=\frac{v_B-u_B}{t_B}\\ =\frac{0-20}{1.0}\\ =-20.0{\text{m/s}}^{\text{2}}\end{array}$

  $\left|a_B\right|=20.0{\text{m/s}}^{\text{2}}$

C.

  $\begin{array}{c}{a}_C=\frac{v_C-u_C}{t_C}\\ =\frac{2.0\times {10}^{-2}-0.4\times {10}^{-2}}{0.50}\\ =0.032{\text{m/s}}^{\text{2}}\end{array}$

D.

  $\begin{array}{c}{a}_D=\frac{v_D-u_D}{t_D}\\ =\frac{4.3-0}{0.40}\\ =10.75{\text{m/s}}^{\text{2}}\end{array}$

E.

  $\begin{array}{c}{a}_E=\frac{v_E-u_E}{t_E}\\ =\frac{1.0-2.0}{8.3}\\ =-0.12{\text{m/s}}^{\text{2}}\end{array}$

  $\left|a_E\right|=0.12{\text{m/s}}^{\text{2}}$

Conclusion:

Order of magnitude of acceleration from least to greatest is as follows:

  $C<E<A<D<B$