Reaction Time to Colors
Unit 1 HW 16.2: Acceleration 2
Name: Date: Period:
UNIT I Acceleration 2
The problem v vs t graph
Solution
1. A poorly tuned Yugo can accelerate from rest to a speed of 28 m/s in 20 s.
a) What is the average acceleration of the car?
b) What distance does it travel in this time?
a = (28-0m/s)/20s = 1.4 m/s/s x = 1/2at2 = ½(1.4)(20)2= 280m
2. At t = 0 a car has a speed of 30 m/s. After 6 s, its speed is 15 m/s.
What is its average acceleration during this time interval?
a = (15-30m/s)/6s = -2.5m/s/s
3. A bear spies some honey and takes off from rest, accelerating at a rate of 2.0 m/s2.
If the honey is 10 m away, how fast will his snout be going at the moment of ecstasy? Hint: Solve for time to go 10m then use time to find velocity
10m = ½(2)t2 10m = t2 = t = 3.16 s. v = at = (2m/s/s)(3.16s) = 6.32 m/s
4. A bus moving at 20 m/s (t = 0) slows at a rate of 4 m/s each second.
a) How long does it take the bus to stop?
b) How far does it travel while braking?
v = at => v/a = t = (20m/s)/(4m/s/s) = 5 s x = vt+1/2at2 = (20m/s)(5s) + ½(-4 m/s/s/)(5s)2 = 100m – 50m = 50 m
5. A car whose initial speed is 30 m/s slows uniformly to 10 m/s in 5 seconds.
a) Determine the acceleration of the car.
v = vi+at = 10m/s = 30 m/s + a(5s) => 10-30 = 5a => = (-10m/s)/(5s) = -4 m/s/s
6. A dog runs down his driveway with an initial speed of 5 m/s for 8 s, then uniformly increases his speed to 10 m/s in 5 s.
a) What was his acceleration during the 2nd part of the motion?
b) How long is the driveway?
v = vi+at = 10m/s = 5 m/s + a(5s) => 10-5 = 5a => =(5m/s)/(5s) = 1 m/s/s
add the distance from the 1st part to the 2nd part xa = (5m/s)(8s) = 40 m. xb vt+1/2at2 = (5m/s)(5s) + ½(1 m/s/s/)(5s)2 = 25m + 12.5m = 37.5 m. xa+xb =40m + 37.5m=77.5m
7. A physics student skis down a slope, accelerating at a constant 2.0 m/s2. If it takes her 15 s to reach the bottom, what is the length of the slope
x =
Name: Date: Period:
UNIT I Acceleration 2
The problem v vs t graph
Solution
1. A poorly tuned Yugo can accelerate from rest to a speed of 28 m/s in 20 s.
a) What is the average acceleration of the car?
b) What distance does it travel in this time?
a = (28-0m/s)/20s = 1.4 m/s/s x = 1/2at2 = ½(1.4)(20)2= 280m
2. At t = 0 a car has a speed of 30 m/s. After 6 s, its speed is 15 m/s.
What is its average acceleration during this time interval?
a = (15-30m/s)/6s = -2.5m/s/s
3. A bear spies some honey and takes off from rest, accelerating at a rate of 2.0 m/s2.
If the honey is 10 m away, how fast will his snout be going at the moment of ecstasy? Hint: Solve for time to go 10m then use time to find velocity
10m = ½(2)t2 10m = t2 = t = 3.16 s. v = at = (2m/s/s)(3.16s) = 6.32 m/s
4. A bus moving at 20 m/s (t = 0) slows at a rate of 4 m/s each second.
a) How long does it take the bus to stop?
b) How far does it travel while braking?
v = at => v/a = t = (20m/s)/(4m/s/s) = 5 s x = vt+1/2at2 = (20m/s)(5s) + ½(-4 m/s/s/)(5s)2 = 100m – 50m = 50 m
5. A car whose initial speed is 30 m/s slows uniformly to 10 m/s in 5 seconds.
a) Determine the acceleration of the car.
v = vi+at = 10m/s = 30 m/s + a(5s) => 10-30 = 5a => = (-10m/s)/(5s) = -4 m/s/s
6. A dog runs down his driveway with an initial speed of 5 m/s for 8 s, then uniformly increases his speed to 10 m/s in 5 s.
a) What was his acceleration during the 2nd part of the motion?
b) How long is the driveway?
v = vi+at = 10m/s = 5 m/s + a(5s) => 10-5 = 5a => =(5m/s)/(5s) = 1 m/s/s
add the distance from the 1st part to the 2nd part xa = (5m/s)(8s) = 40 m. xb vt+1/2at2 = (5m/s)(5s) + ½(1 m/s/s/)(5s)2 = 25m + 12.5m = 37.5 m. xa+xb =40m + 37.5m=77.5m
7. A physics student skis down a slope, accelerating at a constant 2.0 m/s2. If it takes her 15 s to reach the bottom, what is the length of the slope
x =