1.)A particle located initially at -6.00 m accelerates from rest to a velocity o
ID: 1329287 • Letter: 1
Question
1.)A particle located initially at -6.00 m accelerates from rest to a velocity of 2.70 m/s in a time interval of 1.30 s.
1a.)Calculate the acceleration of the particle.
1b)Calculate the distance the particle moves during the 1.30 s time interval.
2.A car is traveling at a constant velocity of 21 m/s (about 45 mph). The driver sees a fallen tree lying across the road and hits the brakes after a delay (reaction time) of 245 ms. (1 ms = 10-3 s.)
2a.)Calculate the distance the car travels during this 245 ms reaction time period.
2b)Once the brakes are applied, the car slows to a stop with an acceleration of -5.29 m/s2. Calculate the distance the car travels while the brakes are applied.
Explanation / Answer
initial velocity = vo = 0
final velocity = v1 = 2.7 m/s
initial position = xo = -6 m
time taken = t = 1.3 s
1a)
a = (v1-vo)/t = (2.7-0)/1.3 = 2.08 m/s^2
1b)
x-xo = vo*t + 0.5*a*t^2
x-xo = 0 + 0.5*2.08*1.3^2
displacement = x-xo = 1.76 m <<------answer
+++
2)
intial velocity vo = 21 m/s
reaction time = t1 = 0.245 s
x1 = vo*t = 21*0.245 = 5.145 m
2b)
x2 = (v1^2-vo^2)/2*a
x2 = (0^2-21^2)/(2*-5.29)
x2 = 41.7 m <<---------answer
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