A car, moving along a straight stretch of high- way, begins to accelerate at 0.0377 m/s2. It takes the car 34.3 s to cover 1 km.
How fast was the car going when it first began to accelerate?
Answer in units of m/s.
d = Vo*t + 0.5*a*t^2 = 1000 m.
Vo*34.3 + 0.5*0.0377*34.3^2 = 1000 m.
34.3Vo + 22.2 = 1000
34.3Vo = 978.
Vo = 28.5 m/s.
To find the initial speed of the car when it first began to accelerate, we can use the equation of motion:
\(s = ut + \frac{1}{2}at^2\)
Where:
s = distance covered (1 km = 1000 m)
u = initial speed of the car
t = time taken (34.3 s)
a = acceleration (0.0377 m/s^2)
Given that s = 1000 m, t = 34.3 s, and a = 0.0377 m/s^2, we can rearrange the equation to solve for u:
\(u = \frac{s - \frac{1}{2}at^2}{t}\)
Substituting the given values, we have:
\(u = \frac{1000 - \frac{1}{2}(0.0377)(34.3^2)}{34.3}\)
Simplifying this expression will give us the initial speed of the car.