Aaron's car is moving at speed of 30 m/s
His reaction time is given as 0.7 s
but when he is tired the reaction time is doubled
Now we need to find the distance covered by his car when he is tired during the time when he react to apply brakes
So here since during this time speed is given as constant so we can say that distance covered can be product of speed and time
So here we can use
So the car will move to 42 m during the time when he apply brakes
Refer to the attachment
Answer:
The particle’s velocity is -16.9 m/s.
Explanation:
Given that,
Initial velocity of particle in negative x direction= 4.91 m/s
Time = 12.9 s
Final velocity of particle in positive x direction= 7.12 m/s
Before 12.4 sec,
Velocity of particle in negative x direction= 5.32 m/s
We need to calculate the acceleration
Using equation of motion
Where, v = final velocity
u = initial velocity
t = time
Put the value into the equation
We need to calculate the initial speed of the particle
Using equation of motion again
Put the value into the formula
Hence, The particle’s velocity is -16.9 m/s.
Answer:
b) 472HZ, 408HZ
Explanation:
To find the frequencies perceived when the bus approaches and the train departs, you use the Doppler's effect formula for both cases:
fo: frequency of the source = 440Hz
vs: speed of sound = 343m/s
vo: speed of the observer = 0m/s (at rest)
v: sped of the train
f: frequency perceived when the train leaves us.
f': frequency when the train is getTing closer.
Thus, by doing f and f' the subjects of the formulas and replacing the values of v, vo, vs and fo you obtain:
hence, the frequencies for before and after tha train has past are
b) 472HZ, 408HZ