Use the worked example above to help you solve this problem. A car traveling at

Question

Use the worked example above to help you solve this problem. A car traveling at a constant speed of 26.1 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 3.92 m/s2 (a) How long does it take the trooper to overtake the speeding car? (b) How fast is the trooper going at that time? m/s EXERCISE HINTS: GETTING STARTED I'M STUCK! A motorist with an expired license tag travels at a constant speed of 21.4 m/s down a street, and a policeman on a motorcycle, taking another 4.89 s to finish his donut, gives chase at an acceleration of 2.03 m/s2 (a) Find the time required to catch the car. (b) Find the distance the trooper travels while overtaking the motorist.

Explanation / Answer

1)

a) let both the car and trooper travel a distance denoted by s (say)

we know that if time taken by the car is (t+1) seconds, then the time taken by the trooper is (t) seconds

distance travelled by car:

s=26.1*(t+1)

distance travelled by trooper:

s= 0*t + 0.5*3.92*t2

equating the value of s from both the equations,

26.1*(t+1) = 0.5*3.92*t2

1.96t2 - 26.1t - 26.1 =0

solving we get

t=14.25 seconds and -0.934 (this value is neglected)

b) using, v = u+at

v= 0+3.92*14.25 = 55.86 m/s

2) similar as the previous problem

a) let both the motorist and police travel a distance denoted by s (say)

we know that if time taken by the motorist is (t+4.89) seconds, then the time taken by the police is (t) seconds

distance travelled by motorist:

s=21.4*(t+4.89)

distance travelled by trooper:

s= 0*t + 0.5*2.03*t2

equating the value of s from both the equations,

21.4*(t+1) = 0.5*2.03*t2

1.015t2 - 21.4t - 21.4 =0

solving we get

t=22.040 seconds and -0.956 (this value is neglected)

b) s=21.4*(t+4.89) = 21.4*(22.040+4.89) = 576.302 metres

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