A particle's position as a function of time t is given by r⃗ =(5.0t+6.0t2)mi^+(7.0−3.0t3)mj^.
2 answers:
The particle's displacement vector has an angle of -65° relative to the positive x-axis.
Additional Details
Acceleration signifies the change in velocity over time.
a = acceleration ( m/s² )
v = final velocity ( m/s )
u = initial velocity ( m/s )
t = time taken ( s )
d = distance ( m )
Now, let's solve the problem!
Data provided:
To find the particle displacement, we need to subtract the two position vectors:
At t = 5.0 seconds,
Further Insights
- Runner's Velocity:
- Kinetic Energy:
- Acceleration:
- Speed of a Car:
Answer Overview
Level: High School
Field: Physics
Topic: Kinematics
Key Terms: Velocity, Driver, Car, Deceleration, Acceleration, Obstacle, Speed, Time, Rate
The particle's position is described as
At t=5 seconds, the position vector reads as
The reference vector when t = 5 seconds is
Let θ denote the angle between the two vectors.
As per definition,
|r₁| = √[175²+(-368)²] = 407.4911
|r₂| = 7
Thus,
θ = cos⁻¹ -2576/(407.4911*7) = cos⁻¹ -0.9031 = 154.57°
Final Result: 154.57°
At t=5 seconds, the position vector reads as
The reference vector when t = 5 seconds is
Let θ denote the angle between the two vectors.
As per definition,
|r₁| = √[175²+(-368)²] = 407.4911
|r₂| = 7
Thus,
θ = cos⁻¹ -2576/(407.4911*7) = cos⁻¹ -0.9031 = 154.57°
Final Result: 154.57°
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