Answer:
a. 2v₀/a b. 2v₀/a
Explanation:
a. Since you are moving with a constant velocity v₀, the distance, s you cover in time = t max is s = v₀t.
Since the dragster starts from rest with an acceleration, a, using
s' = ut + 1/2at² where u = 0 and s' = distance moved by dragster
s' = 0t + 1/2at²
s' = 1/2at²
Since the distance moved by me and the dragster must be the same,
s = s'
v₀t. = 1/2at²
v₀t. - 1/2at² = 0
t(v₀ - 1/2at) = 0
t= 0 or v₀ - 1/2at = 0
t= 0 or v₀ = 1/2at
t= 0 or t = 2v₀/a
So the maximum time tmax = 2v₀/a
b. Since the distance covered by me to meet the dragster is s = v₀t in time, t = tmax which is also my distance from the dragster when it started. So, my distance from the dragster when it started is s = v₀(2v₀/a)
= 2v₀/a
Answer:
Explanation:
Yes , their displacement may be equal .
Suppose the displacement is AB where A is starting point and B is end point .
The car is covering the distance AB by going from A to B on straight line . On the other hand plane goes from A to C , then from C to D and then from D to B . In this way plane reaches B from A on a different path which is longer than path of the car . In the second case also displacement of plane is AB . In the second case distance covered is longer but displacement is same that is AB .
I believe it’s A, i could be wrong tho 3
Answer:
150J
Explanation:
Formula : <u>Work</u><u> </u><u>done</u>
Force x distance
work done = force x distance
Distance should be measured in meters
300÷100=3m
work done = 450 x 3
=150J
The most common value for the speed of light is 3*10^8 meters/second.
A more accurate number is <span>299 792 458 m/second, but that number is hardly ever used.</span>
