Answer:
v = 49.69 km/hr
Explanation:
velcoity of truck with respect to car is given as = velocity of truck - velocity of car
velcoity of truck with respect to car is given as = (-54 sin(28)i +54cos(28)j - 91km/h
= -24.35i - 43.32j
magnitude of velocity is

v = 49.69 km/hr
the direction is
![tan^{-}[\frac{ 43.32}{24.35}]](https://tex.z-dn.net/?f=tan%5E%7B-%7D%5B%5Cfrac%7B%2043.32%7D%7B24.35%7D%5D)
= 60.65 degree towrd south of west
Answer:
<u><em>note:</em></u>
<u><em>solution is attached due to error in mathematical equation. find the attachment</em></u>
Answer:
12.0 meters
Explanation:
Given:
v₀ = 0 m/s
a₁ = 0.281 m/s²
t₁ = 5.44 s
a₂ = 1.43 m/s²
t₂ = 2.42 s
Find: x
First, find the velocity reached at the end of the first acceleration.
v = at + v₀
v = (0.281 m/s²) (5.44 s) + 0 m/s
v = 1.53 m/s
Next, find the position reached at the end of the first acceleration.
x = x₀ + v₀ t + ½ at²
x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²
x = 4.16 m
Finally, find the position reached at the end of the second acceleration.
x = x₀ + v₀ t + ½ at²
x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²
x = 12.0 m