Question #137492
1.A car is traveling at 90 km/hr. In coming to a stop, it has an acceleration of –8.0 m/s2 . If the reaction
time of the driver is 2.5 s, how far will the car travel before coming to a rest?

2.A speeder passes a parked police car at 35.0 m/s. The police car starts from rest with a uniform
acceleration of 2.60 m/s2
a) How much time passes before the speeder is overtaken by the police car?
b) How far does the speeder get before being overtaken by the police car?

Runner A is initially 4.5 km west of a flagpole and is running with a constant velocity of 5.0 km/h due east.
Runner B is initially 6.5 km east of the flagpole and is running with a constant velocity of 4.5 km/h due west.
What will be the distance of the two runners from the flagpole when their paths cross?
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1
Expert's answer
2020-10-12T07:49:18-0400

1) s=v2v022a=022522(8)=39.1(m)s=\frac{v^2-v_0^2}{2a}=\frac{0^2-25^2}{2\cdot (-8)}=39.1(m)


2) s1=v0t,s2=gt22s1=s2v0t=gt22t=2v0/g=235/9.81=7.1(s)s_1=v_0t,s_2=\frac{gt^2}{2}\to s_1=s_2\to v_0t=\frac{gt^2}{2}\to t=2v_0/g=2\cdot35/9.81=7.1(s)


s1=357.1250(m)s_1=35\cdot7.1\approx250(m)


3) x1+v1t=x2+v2tx_1+v_1t=x_2+v_2t\to


=4.5+5t=6.54.5tt=1.158(hours)=-4.5+5t=6.5-4.5t\to t=1.158(hours)


s=6.54.51.158=1.29(km)s=6.5-4.5\cdot1.158=1.29(km)














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