Chapter 3.2, Problem 26PP

To determine

To Calculate: The initial velocity of the bike.

Answer to Problem 26PP

  $0.945\text{m/s}$

Explanation of Solution

Given info:

A bike rider reaches a velocity of $7.5\text{ m/s}$ north in a time of 4.5s by travelling at a constant acceleration. During this period, the displacement of bike is 19m north.

Formula used:

Final velocity with constant acceleration can be obtained by:

  $v_{final}=v_{initial}+a\Delta t$

Where, $\Delta t$ is the time taken, a is acceleration and v represents velocity.

The distance traveled by an object with a constant acceleration can be obtained by:

  $d_{final}=d_{initial}+v_{initial}\Delta t+\frac{1}{2}a\Delta t^2$

Here, d represents displacement, v represents velocity, t represents time and a is the acceleration.

Calculation:

First, find acceleration in terms of initial velocity:

  $\begin{array}{c}{v}_{final}=v_{initial}+a\Delta t\\ 7.5=v_{initial}+a\left(4.5\right)\\ a=\frac{7.5-v_{initial}}{4.5}\mathrm{...........}\left(1\right)\end{array}$

Now, assume the initial position of the car to be 0, that is:

  $d_{initial}=0\text{m}$

Then, the distance travelled by the car in 4.5s can be written as:

  $\begin{array}{c}{d}_{final}=0+\left(v_{initial}\right)\left(4.5\right)+\frac{1}{2}\left(a\right){\left(4.5\right)}^2\\ 19=4.5v_{initial}+10.125a\end{array}$

Substitute equation (1) in above equation and solve.

  $\begin{array}{c}19=4.5v_{initial}+10.125\left(\frac{7.5-v_{initial}}{4.5}\right)\\ 19=4.5v_{initial}+16.875-2.25v_{initial}\\ v_{initial}=\frac{2.125}{2.25}\\ =0.945\text{m/s}\end{array}$

Conclusion:

Thus, the initial velocity of the given bike is $0.945\text{m/s}$ .