A particle moving along a straight line is subjected to a deceleration a=(-2v3) m/s2, where v is in m/s. If it has a velocity v=8m/s and a position s=10m whent=0, determine its velocity and position when t=4s.
\(a=-2 v^{3}\)
\(\frac{d v}{d t}=-2 v^{3}\)
Now
\(\int \frac{d v}{-v^{3}}=\int 2 d t\)
Integrate
\(\frac{1}{2 v^{2}}=2 t+C\)
\(v^{2}=\frac{1}{4 t+2 C}\)
At \(t=0, \quad v=8\)
\(\frac{1}{2(64)}=0+C \quad \Rightarrow C=\frac{1}{128}\)
\(v^{2}=\frac{1}{4 t+\frac{1}{64}}\)
\(v^{2}=\frac{64}{256 t+1}\)
\(v=\frac{8}{\sqrt{256 t+1}}\)
\(v_{t-4}=\frac{8}{\sqrt{1024+1}}\)
\(v_{t-4}=0.25 \mathrm{~ms}\)
\(\frac{d s}{d t}=\frac{8}{\sqrt{256 t+1}}\)
\(\int d s=\int \frac{8}{\sqrt{256 t+1}} d t\)
\(s=\frac{8(2 \sqrt{256 t+1})}{256}+k\)
\(s=\frac{\sqrt{256 t+1}}{16}+k\)
\(A \cdot t=0, s=10\)
\(10=\frac{1}{16}+k\)
\(k=10-\frac{1}{16}\)
\(k=\frac{159}{16}\)
\(s=\frac{\sqrt{256 t+1}+159}{16}\)
\(s_{t-4}=\frac{\sqrt{1025}+159}{16}\)
\(s_{t-4}=11.94 m\)
A particle travels along a straight line with a velocity v=(12-3t2) m/s, where t is in seconds. When t=1s, the particle is located 10m to the left of the origin. Determine the acceleration when t=4s, the displacement from t=0 to t=10s, the distance the particle travels during this time period.
The velocity of a particle moving along a straight line is given by v = 0.2s1/2 m/s where the position s is in meters. At t = 0 the particle has a velocity v0 = 3 m/s. Determine the time when the particle’s velocity reaches 15 m/s and the corresponding acceleration. Ans: 600 s, a = 0.02 m/s2
12-17. and t 0. If it is subjected to a deceleration of a--k where k is a constant, determine its velocity and position as functions of time. A particle is moving with a velocity of vo when s 0
12-15. A particle is moving with a velocity of vo when s-0 and t 0. If it is subjected to a deceleration of a-kv3, where k is a constant, determine its velocity and position as functions of time,
2. A particle is traveling along a straight line and the acceleration is a (0.5s*) m/s2, where s is in meters. Whent-0,s-0 and v- 0, please find the particle velocity when s 2 m.
3. The particle travels along a straight line with a velocity (22 - 5t) m/s, where t is in seconds. If s 10 m whent0, determine the following: a. The position of the particle when t-4s b. The total distance traveled during the time interval from t-0 to 4 s c. The acceleration when t 2s
the velocity of a particle traveling in a straight line is given by v=(6t-3t^2)m/s, where t is in seconds, if s=0 when t= 0. determine the particles deceleration and position when t=3s. how far has the particle traveled during the 3s time interval and what is its average speed?
A particle is moving along a straight path such that the acceleration a = (3v2-2) m/s2, where v is in m/s. If v = 15 m/s when s = 0 and t = 0, please determine the particle’s position, velocity, and acceleration as functions of time.
A particle is moving along a straight path such that the acceleration a = (3v-2) m/s2, where v is in m/s. If v = 15 m/s when s = 0 and t = 0, please determine the particle’s position, velocity, and acceleration as functions of time.
A particle is moving along a straight line with an initial velocity of 6 m/s when it is subjected to a deceleration of a = (-1.5012) m/s², where vis in m/s. Determine how far it travels before it stops. How much time does this take?