Motion in one dimension The position of a certain particle depends on time accor

Question

Motion in one dimension The position of a certain particle depends on time according to the equation x(t) = t^2 - 5t + 1, where x is in meters and t is in seconds. Find the displacement and average velocity for the interval 3s lessthanorequalto t lessthanorequalto 4s: Obtain the instantaneous velocity and instantaneous acceleration for any time t. At what time is the particle momentarily at rest? Where is the particle when momentarily at rest? At what times is the particle found at the origin of the x axis? Ans. Delta x = 2m v = 2t - 5 (m/s) a = 2 (m/s^2) t = 2.5 s x = -5.25 m t_1, 2 = 0.21 s & 4.79 s The position of an object is related to time by x(t) = 8t^2 - 6t + 4 (x in meters time in seconds). Find the instantaneous velocity and acceleration as function of time. Ans. v = 16t - 6 (m/s) a = 16 (m/s^2) At t = 5s, an object at x = 3m has a velocity of 5 m/s. At t = 8s, it is at x = 9m and its velocity is -1 m/s. Find the average velocity for this time interval. find the average acceleration for this time interval. Ans. v^bar = 2m/s a^bar = -2m/s^2 A car starting at x = 50m accelerates from rest at a constant rate of 8 m/s^2. How fast is it going after 10 s? Where is the car at the end of the 10s time interval? What is its average velocity for the interval o lessthanorequalto t lessthanorequalto 10s? Ans. v = 80 m/s x = 450 m v^bar = 8m/s An object with constant acceleration has a velocity of 10m/s when it is at x = 6m, and of 15m/s when it is at x = 10m. What is its acceleration? Ans. a = 15.625 m/s^2 A bus accelerates at 1.5 m/s^2 from rest for 12 s. It then travels at constant velocity for 25 s, after which it slows to a stop with an acceleration of -1.5 m/s How far does the bus travel? What is its average velocity? Ans: d = 666 m v_average = 13.6 m/s A ball is thrown upward with an initial velocity of 20 m/s. How long does it take the ball to reach maximum height? What is the greatest height reached by the ball? How long does it take the ball to return to the ground? What is the velocity of the ball just before it hits the ground?

Explanation / Answer

1. Given : x(t)= t2-5t+1

(a)

dx/dt = 2t-5

3s<t<4s

dx/dt = 2-5 = -3

Displacement is :

at t= 3s , x = 9-15+1 = -5

at t= 4s , x = 16-20+1 =-3

Displacement = -5-(-3) = -2 i.e. x = 2m

(b) instantaneous velocity at any time

dx/dt = 2t-5 (m/s)

Instantaneous acceleration

d2x/dt2= 2 m/s2

(c) at rest , dx/dt = 2t-5= 0

t = 5/2 = 2.5 s

x = (2.5)2-5(2.5)+1 = -5.25 m

(d) at origin , x(t)= 0

t2-5t+1= 0

Solving equation we get , t= 0.21s and 4.79 s

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