Chapter 3, Problem 71GP

(a)

To determine

The minimum speed of the car for horizontal ramp.

Answer to Problem 71GP

The minimum speed of the car to cross the horizontal distance is $36.36\text{m}$ .

Explanation of Solution

Given:

Height of the ramp, $h=1.5\text{m}$ .

Horizontal distance for the jump of the car, $x=20\text{m}$

Formula Used:

The expression to calculate the time for crossing the car using second law of motion is,

$h=ut+\frac{1}{2}g{t}^2$    … (1)

Where,

  $u$ is the initial speed of the car and its value is zero.

  $t$ is the time taken.

  $g$ is the gravitational acceleration.

The expression to calculate the minimum speed of the car to cross the horizontal distance is,

$v=\frac{x}{t}$    … (2)

Calculation:

Substitute all the values in equation (1).

$\begin{array}{c}1.5\text{m}=\left(0\text{m}/\text{s}\right)t+\frac{1}{2}\left(9.8\text{m}/{\text{s}}^2\right)t^2\\ t=\sqrt{\frac{2\left(1.5\text{m}\right)}{9.8\text{m}/{\text{s}}^2}}\\ =0.55\text{s}\end{array}$

Substitute all the values in equation (2).

$\begin{array}{c}v=\frac{20\text{m}}{0.55\text{s}}\\ =36.36\text{m}\end{array}$

Conclusion:

Thus, the minimum speed of the car to cross the horizontal distance is $36.36\text{m}$ .

(b)

To determine

The minimum speed of the car for tilted ramp.

Answer to Problem 71GP

The minimum speed of the car to cross the tilted ramp is $20.05\text{m}/\text{s}$ .

Explanation of Solution

Given:

Angle of tilt of the ramp, $\theta =10°$ .

Formula Used:

Write the expression to calculate the take taken when the car is tilt at an angle.

$t=\frac{x}{v_0\cos \theta }$

Where,

  $x$ is the distance covered.

  $\theta$ is the inclination angle.

Consider the second law of motion.

$h=y_0+v_{x}t+\frac{1}{2}{a}_y{t}^2$

Where,

  $h$ is the height.

  $v_x$ is horizontal velocity component.

  $a_y$ is the acceleration along the direction of the motion.

Substitute the values in the above expression.

$\begin{array}{c}h=0+v_0\sin \theta \cdot \frac{x}{v_0\cos \theta }+\frac{1}{2}\left(-g\right){\left(\frac{x}{v_0\cos \theta }\right)}^2\\ h=x\tan \theta -\frac{1}{2}g{\left(\frac{x}{v_0\cos \theta }\right)}^2\\ x\tan \theta -h=\frac{1}{2}g{\left(\frac{x}{v_0\cos \theta }\right)}^2\end{array}$

On solving, the expression to calculate the minimum speed of the car to cross the tilted ramp is,

$v_0=\sqrt{\frac{g{x}^2}{2\left(x\tan \theta +h\right){\cos }^2\theta }}$

Calculation:

Substitute all the values in the above expression.

$\begin{array}{c}{v}_0=\sqrt{\frac{\left(9.8\text{m}/{\text{s}}^2\right){\left(20\text{m}\right)}^2}{2\left(\left(20\text{m}\right)\left(\tan 10°\right)+1.5\text{m}\right)\left({\cos }^{2}10°\right)}}\\ =20.05\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the minimum speed of the car to cross the tilted ramp is $20.05\text{m}/\text{s}$ .