A )
t 1 = 2 h, t 2 = 6 h
Δ t = t 2 - t 1 = 6 - 2 = 4 h
54 = 50 + a Δ t
54 = 50 + 4 a
4 a = 54 - 4
4 a = 4
a = 4 : 4
a = 1 km/h²
v o = 48 km/h
An equation that can be used to describe the velocity of the car at the different times is:
v = 48 + t
B ) The graph is in the attachment.
The magnetic field of the wire will be directed towards west. Using right thumb rule one can get the direction of field lines.
Answer:
yes
Explanation:
I think so because it not mention in the law
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
