Problem#1
A car is approaching a hill at 30.0 m/s when its engine suddenly fails just at the bottom of the hill. The car moves with a constant acceleration of -2.00 m/s2 while coasting up the hill. (a) Write equations for the position along the slope and for the velocity as functions of time, taking x = 0 at the bottom of the hill, where vi = 30.0 m/s. (b) Determine the maximum distance the car rolls up the hill.
Answer:
(a) Take t0 = 0 at the bottom of the hill where x0 = 0 , v0 = 30.0 m/s, a = −2.00 m/s2. Use these values in the general equation
x = x0 + v0t + ½ at2
to find
x = 0 + (30 m/s)t + ½ (−2.00 m/s2)t2 or
x = (30.0t – t2) m
To find an equation for the velocity, use
v = v0 + at
v = 30 m/s + (−2.00 m/s2)t
v = (30.0 – 2.0t) m/s
(b) The distance of travel x becomes a maximum, xmax , when v = 0 (turning point in the motion). Use the expressions found in part (a) for v to find the value of t when x has its maximum value:
From v = (30.0 – 2.0t) m/s, v = 0 when t = 15.0 s. Then
xmaks = (30.0t – t2) m = (30.0 m/s)(15.0 s) – (15.0 s)2 = 225 m
Problem#2
A truck on a straight road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 20.0 m/s. Then the truck travels for 20.0 s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.00 s. (a) How long is the truck in motion? (b) What is the average velocity of the truck for the motion described?
Answer:
(a) The time it takes the truck to reach 20 0. m s is found from
v = v0 + at
Solving for t yields
t = (v – v0)/a = (20 m/s – 0)/(2.00 m/s2)
t = 10.0 s
The total time is thus
10.0 s + 20.0 s + 5.00 s = 35.0 s
(b) The average velocity is the total distance traveled divided by the total time taken. The distance traveled during the first 10.0 s is
x1 = vavgt = (0 + 20.0 m/s)(10.0 s)/2 = 100 m
With a being 0 for this interval, the distance traveled during the next 20.0 s is
x2 = v0t + ½ at2
x = (20.0 m/s)(20.0 s) + 0 = 400 m
The distance traveled in the last 5.00 s is
x3 = vavgt = (0 + 20.0 m/s)(5.00 s)/2 = 50.0 m
The total distance x1 + x2 + x3 = 100 m + 400 m + 50.0 m = 550 m , and the average velocity is given by
vavg = x/t = 550 m/35.0 s = 15.7 m/s
Problem#3
A glider on an air track carries a flag of length l through a stationary photogate, which measures the time interval ∆td during which the flag blocks a beam of infrared light passing across the photogate. The ratio vd = L/∆td is the average velocity of the glider over this part of its motion. Suppose the glider moves with constant acceleration. (a) Argue for or against the idea that vd is equal to the instantaneous velocity of the glider when it is halfway through the photogate in space. (b) Argue for or against the idea that vd is equal to the instantaneous velocity of the glider when it is halfway through the photogate in time.
Answer:
Let the glider enter the photogate with velocity vi and move with constant acceleration a. For its motion from entry to exit,
x – x0 = v0xt + ½ axt2
l = 0 + v0 ∆td + ½ a∆td2 = vd∆td
vd = v0 + ½ a∆td
(a) The speed halfway through the photogate in space is given by
vhs2 = v02 + 2a(l/2) = v02 + avd∆td
vhs = [v02 + avd∆td]1/2
and this is not equal to vd unless a = 0.
(b) The speed halfway through the photogate in time is given by
vhs2 = v02 + a(∆td/2)
and this is equal to vd as determined above.
Problem#4
Speedy Sue, driving at 30.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 155 m ahead traveling at 5.00 m/s. Sue applies her brakes but can accelerate only at -2.00 m/s2 because the road is wet. Will there be a collision? If yes, determine how far into the tunnel and at what time the collision occurs. If no, determine the distance of closest approach between Sue’s car and the van.
Answer:
Take the original point to be when Sue notices the van. Choose the origin of the x-axis at Sue’s car. For her we have x0s = 0, v0s = 30.0 m/s, as = −2.00 m/s2 so her position is given by
xs(t) – x0s = v0st + ½ ast2
xs(t) = (30.0 m/s)t + ½ (−2.00 m/s2)t2
For the van,
x0v = 155 m, v0v = 5.00 m/s, av = 0 and
xv(t) – x0v = v0vt + ½ avt2
xs(t) = 155 m + (5.00 m/s)t + 0
To test for a collision, we look for an instant tc when both are at the same place:
30.0tc – tc2 = 155 + 5.00tc
tc2 – 25.0tc + 155 = 0
From the quadratic formula
tc = [+25.0 ± {(25.0)2 – 4(1)(155)}1/2]/2
tc = 13.6 s or 11.4 s
The smaller value is the collision time. (The larger value tells when the van would pull ahead again if the vehicles could move through each other). The wreck happens at position
155 m + (5.00 m/s)(11.4 s) = 212 m
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