A car, moving along a straight stretch of highway, begins to accelerate at 0.0261 m/s2. It takes the car 48.6 s to cover 1 km.
How fast was the car going when it first
began to accelerate?
Answer in units of m/s
Couldn't find a formula for this :(
d=vi*t+1/2 a t^2
1=vi*48.6+.01305(48.6)^2
solve for vi
To determine the initial velocity of the car when it first began to accelerate, we can use the following kinematic equation:
s = ut + (1/2)at^2
Where:
- s is the distance covered (1 km = 1000 m)
- u is the initial velocity
- t is the time taken (48.6 s)
- a is the acceleration (0.0261 m/s^2)
Using this equation, we can rearrange it to solve for u:
u = (s - (1/2)at^2) / t
Plugging in the known values:
u = (1000 m - (1/2)(0.0261 m/s^2)(48.6 s)^2) / 48.6 s
Simplifying the equation:
u = (1000 m - 59.375 m) / 48.6 s
u = 940.625 m / 48.6 s
u ≈ 19.35 m/s
Therefore, the car was initially going at approximately 19.35 m/s when it first began to accelerate.