A police car, initially at rest, is passed by a black Ford Mustang traveling at

Question

A police car, initially at rest, is passed by a black Ford Mustang traveling at a constant 130 km/h. The police car, maintaining a constant acceleration, catches up to the speeding Mustang in 750 m. Let t = 0 be the instant the Mustang passes the resting police car. Ignore any reaction time from the police officer. so assume the police car begins accelerating at constant rate at t = 0. Determine the time at which the police car catches up to the speeding Mustang. Determine the required police car acceleration, and the speed of the police car at the catch-up point. Model: (a) State what simplifying assumptions you are making to solve this problem, and how each assumption helps you solve the problem. Visualize: (b) Draw motion diagrams for both care. (c) Draw a second visualization that is appropriate for how you are going to solve the problem. For example: If you arc going to solve this problem graphically, then drawing a qualitative position vs, time graph and/or velocity vs, time graph probably makes most sense. If you are going to solve this problem mathematically with the kinematic equations, a pictorial representation like the one in Example 2.11 in the textbook probably makes most sense. Solve: (d) Solve the problem, symbolically first, then inserting values. (This requires that you assign variables to each of the known and unknown values. See Example 2.11 in the textbook for guidance.) Assess: (e) At the catch-up point, how does the speed of the police car compare to the speed of the Mustang? Does this make sense? (f) Quantitatively graph position vs, time for both cars. Use this graph to verify your results.

Explanation / Answer

speed of mustang Vm = 130 km/h

distance = 750 m = 0.75 km

so time t in which police car catches speed of 130 km/h starting from rest in distance of 750 m

we have 3 equations of motion,

S= u*t + 1/2*a*t2

v2-u2= 2*a*s

v= u + a*t

here u = inital velocity,

v = final velocity

a= acceleration

t= time

s= distance

so in our case

using

v2-u2= 2*a*s

1302 - 0 = 2 * a * 0.750

hence a = 11266.67 km/h2

and time t will be =

v= u + at

130 =0 + 11266.67 t

hence t= 0.011 Hrs = 41.53 sec

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