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notka56 [123]
4 years ago
15

A rocket-driven sled running on a straight level track is used to investigate the physiological effects of large accelerations o

n humans. One such sled can attain a speed of 1000 rniles/hr in 1.8 sec starting from rest. (1 mile = 5280 ft) a. Assume the acceleration is constant and compare it to g. (g=9.8 m/s2 = 32 ft/ s2) b. What is the distance traveled in this time?
Physics
1 answer:
Shtirlitz [24]4 years ago
4 0

Answer:

A. a = 814.815 ft/s^2.

B. Distance, S = 320 ft.

Explanation:

Equations of motion

i. vf = vi + a*t

ii. S = vi*t + 1/2*(a*t)

iii. vf^2 = vi^2 + 2a*S

Given:

vi = 0 ft/s

vf = 1000 miles/hr

t = 1.8 s

g = 32 ft/ s2

Converting miles/hr to ft/s,

1 mile = 5280 ft

Also, 1 hr = 60 mins * 60s

= 3600 s

Therefore, 1000 miles/hr * 5280 ft/1 mile * 1 hr/3600 s

= 1466.7 ft/s.

A.

Using the i. Equation of motion,

1466.7 = 0 + a*1.8

a = 814.815 ft/s^2

Comparing a to g,

a = (814.815/32) * g

= 25.46 g

B.

Using the ii. Equation of motion,

S = 1/2 * (814.815) * (1.8)^2

= 1320 ft.

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A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr
KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s

Time the parachutist falls without friction is 3.19 seconds

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2

Now the initial velocity of the last half height will be the final velocity of the first half height.

v=u+at\\\Rightarrow v=at

Since the height are equal

\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0

t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2

Magnitude of acceleration is -43200 ft/s²

5 0
3 years ago
A student throws a baseball horizontally at 25 meters per second from a cliff 45 meters above the level ground. Approximately ho
Delicious77 [7]
First, calculate how long the ball is in midair. This will depend only on the vertical displacement; once the ball hits the ground, projectile motion is over. Since the ball is thrown horizontally, it originally has no vertical speed. 
t = time vi = initial vertical speed = 0m/s g = gravity = -9.8m/s^2 y = vertical displacement = -45m 
y = .5gt^2 [Basically, in this equation we see how long it takes the ball to fall 45m] -45m = .5 (-9.8m/s^2) * t^2 t = 3.03 s 
Now we know that the ball is midair for 3.03s. Since horizontal speed is constant we can simply use: 
x = horizontal displacement v = horizontal speed = 25m/s t = time = 3.03s 
x = v*t x = 25m/s * 3.03s = 75.76 m Thus, the ball goes about 75 or 76 m from the base of the cliff.
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