A car traveling with an initial velocity of 68 km/h accelerates uniformly at a rate of 1.35 m/s^2 for 4.7s. Find the final speed and displacement of the car during the specified time interval.
Vo = 68km/h = 68,000m/3600s = 18.89 m/s.
V=Vo + a*t = 18.89 + 1.35*4.7 = 25.2m/s
= Final velocity.
D = Vo*t + 0.5*a*t^2
D = 18.89*4.7 + 0.5*1.35*4.7^2=103.7 m.
To find the final speed and displacement of the car during the specified time interval, we can use the formulas of uniformly accelerated motion.
1. Final speed (v):
The formula to calculate the final speed of an object undergoing uniform acceleration is:
v = u + a * t
Where:
- v is the final speed
- u is the initial velocity
- a is the acceleration
- t is the time interval
Using the given values:
u = 68 km/h = 68,000 m / (60 * 60) s (convert km/h to m/s)
a = 1.35 m/s^2
t = 4.7 s
Plugging these values into the formula, we get:
v = (68,000 / (60 * 60)) + (1.35 * 4.7)
Now we can calculate it:
v = 18.89 + 6.33
Therefore, the final speed of the car during the specified time interval is approximately 25.22 m/s.
2. Displacement (s):
The formula to calculate the displacement of an object undergoing uniform acceleration is:
s = u * t + 0.5 * a * t^2
Using the given values:
u = 68 km/h = 68,000 m / (60 * 60) s (convert km/h to m/s)
a = 1.35 m/s^2
t = 4.7 s
Plugging these values into the formula, we get:
s = (68,000 / (60 * 60)) * 4.7 + 0.5 * 1.35 * (4.7)^2
Now we can calculate it:
s = 71.11 + 14.83
Therefore, the displacement of the car during the specified time interval is approximately 85.94 meters.