A car accelerates at 2.25 m/s^2 along a straight road. It passes two marks that are 27.8 m apart at times t=3.80 s and t=5.25 s. What was the car's velocity at t=0?
Let distances, S1 and S2, correspond to the distances at t1=3.8 and t2=5.25 respectively, assuming the distance =0 at t=0.
acceleration, a = 2.25 m/s/s
S1=u*(t1)+(1/2)a(t1)²
S2=u*(t2)+(1/2)a(t2)²
S2-S1=u(t2-t1)+(1/2)a(t2²-t1²)
Since S2-S1, t1,t2 and a are known, u can be isolated and solved for.
To find the car's velocity at t=0, we can use the equation of motion:
\(v = u + at\)
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
The car passes two marks that are 27.8 m apart at times t = 3.80 s and t = 5.25 s.
Let's first find the final velocity at t = 5.25 s. Since we don't know the initial velocity, we can use the equation:
\(v = u + at\)
For t = 5.25 s, v = final velocity, u = initial velocity, a = acceleration, and t = 5.25 s. Plugging in the values, we get:
\(v = u + (2.25 \, \text{m/s}^2) \cdot (5.25 \, \text{s})\)
To simplify, multiply the acceleration and time:
\(v = u + 11.8125 \, \text{m/s}\)
Next, let's find the final velocity at t = 3.80 s. Again, using the equation:
\(v = u + at\)
For t = 3.80 s, v = final velocity, u = initial velocity, a = acceleration, and t = 3.80 s. Plugging in the values, we get:
\(v = u + (2.25 \, \text{m/s}^2) \cdot (3.80 \, \text{s})\)
To simplify, multiply the acceleration and time:
\(v = u + 8.55 \, \text{m/s}\)
Now we have two equations:
Equation 1: \(v = u + 11.8125 \, \text{m/s}\)
Equation 2: \(v = u + 8.55 \, \text{m/s}\)
We can subtract Equation 2 from Equation 1 to eliminate the variable "v":
\(v - v = (u + 11.8125 \, \text{m/s}) - (u + 8.55 \, \text{m/s})\)