Answer :
The first part is pretty straight forward, vₓ=v₀+a(t). Convert final velocity to m/s to get 97.2m/s. v₀=0, a=4. Solving for t yields 24.3 seconds.
For the second part use "vₓ^2(final velocity) = v₀^2(initial velocity) + 2a(Δx). Final velocity is 97.2m/s, initial velocity is 0, acceleration is 4m/s^2. Solving for Δx yields 1181.53 meters.
For the second part use "vₓ^2(final velocity) = v₀^2(initial velocity) + 2a(Δx). Final velocity is 97.2m/s, initial velocity is 0, acceleration is 4m/s^2. Solving for Δx yields 1181.53 meters.
Answer:
The distance traveled by the car during this time is 1.18 km.
Explanation:
Initial speed of the car, u = 0 (at rest)
Final sped of the car, [tex]v=350\ km/h = 97.23\ m/s[/tex]
Acceleration of the car, [tex]a=4\ m/s^2[/tex]
We need to find the distance would the car travel during this time. It can be calculated using third equation of motion. It is given by :
[tex]v^2-u^2=2ad[/tex]
[tex]v^2=2ad[/tex]
[tex]d=\dfrac{v^2}{2a}[/tex]
[tex]d=\dfrac{(97.22\ m/s)^2}{2\times 4\ m/s^2}[/tex]
d = 1181.46 m
or
d = 1.18 km
So, the distance traveled by the car during this time is 1.18 km. Hence, this is the required solution.