Question

OAssigment submit dep 20213641 REMARKS The trooper, traveling about twice as fast as the car, must swerve or apply his brakes strongly to avoid a collisionl This problem can also be solved graphically by plotting position versus time for each vehicle on the same graph. The intersection of the two graphs corresponds to the time and position at which the trooper overtakes the car QUESTION The graphical solution corresponds to finding the intersection of what two types of curves in the tx-plane? (Select all that apply.) O a straight line sloped downward 2 an upward-shaped curve whose slope increases as the displacement x increases O a downward-shaped curve whose slope decreases and then increases as x incr a straight horizontal line U a straight line sloped upward Examine the expression for the position as a function of time for each of the two cars. Determine how the position vs. time graph of each would appear when plotted in the tx-plane. PRACTICE IT Use the worked example above to help you solve this problem. A car traveling at a constant speed of 27.5 m/s passes a trooper hidden behind a billboard, as in the figure. One second after the speeding car passes the billboard, the trooper sets off in chase with a constant acceleration of 3.62 m/s (a) How long does it take the trooper to overtake the speeding car? 15 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. s (b) How fast is the trooper going at that time? 14 66 Your response differs from the correct answer by more than 10%. Double check your calculations. m/s

Answer 1

Red line is x-t curve for car's motion. ( straight line sloped upward)

Blue curve is x-t curve for trooper's motion. (an upward shaped curve whose slope increases as the displacement x

increases )

Practice it:

Velocity of car $v=27.5\,m/s$ (constant)

The car travels a distance $S_1$ in time $t_1=1.0\,s$

$S_1=vt_1=27.5*1=27.5\,m$

Then the car travels a distance $S_2$ in time $t_2$ .

$S_2=vt_2=27.5t_2$

After time $t_1$ , the trooper starts from rest $u=0\,m/s$ with an acceleration $a=3.62\,m/s^2$ . In time $t_2$ trooper travels a distance $S$ and overtakes car.

$S=ut_2+\frac{1}{2}at_2^2=0+\frac{1}{2}*3.62t_2^2=1.81t_2^2$

The distance traveled by trooper in time $t_2$ is equal to total distance covered by car in time $t_1+t_2$

That is $S=S_1+S_2$

$1.81t_2^2=27.5+27.5t_2$

$1.81t_2^2-27.5t_2-27.5=0$

solving above quadratic equation $t_2=16.135\,s$

a) Time taken by trooper to overtake the car is $t_2=16.135\,s\approx 16\,s$

b) Speed of trooper while he is overtaking car is $v=u+at_2=0+3.62*16.135=58.4\,m/s$