Concept - find the average acceleration by finding the velocity at the given time. Differentiate the equation of velocity to find the instantaneous acceleration. Put the values of time to find the value of instantaneous acceleration as shown below.

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suppose the x-velocity vx of the car at any time t is given by the equation:...
A car’s position as a function of time is given by the following equation: x(t) = 5 m/s t + 2.8 m/s2t2– 0.15 m/s3t3. Find the average velocity from 0 to 5 s. Find the instantaneous velocity at 0, 3, and 5 s. Find the average acceleration from 0 to 5 s. Find the instantaneous acceleration at 0, 3, and 5 s. At what POSITIVE time does the car come to rest?
4. A car's position as a function of time is given by the following equation: x(t)-5 m/s t+2.8 m/s2 t-0.15 m/s3 t3. a. Find the average velocity from 0 to 5 s b. Find the instantaneous velocity at 0, 3, and 5s. c. Find the average acceleration from 0 to 5 s. d. Find the instantaneous acceleration at 0, 3, and 5 s. e. At what POSITIVE time does the car come to rest?
EXAMPLE 3 Show that the average velocity of a car over a time interval [t1, tz] is the same as the average of its velocities during the trip. SOLUTION If s(t) is the displacement of the car at time t, then, by definition, the average velocity of the car over the interval is s(t2) - s(tı) As At On the other hand, the average velocity function on the interval is Vave = v(t) dt = s'(t) dt t2 t-taf. “t,'t...
A stock car starts from rest at time t=0 with velocity (m/s) increasing for 3.7 s , according to the function vx=1.4t2+1.1t. Find the average acceleration for this interval
A sports car was tested, and its velocity was given by the equation v=b(1-e-t/c) where b=128.1 m/s and c=13.31 sec. How long did it take the car to reach 400km/hr? What was its average acceleration during the test? What is the car's theoretical speed? Derive an expression for acceleration as a function of time. What was the acceleration of the car when the test started? What was the acceleration of the car when it hit 400km/h?
A dog running in an open field has components of velocity vx = 2.60 m/s and vy = -1.40 m/s at time t1 = 10.9 s . For the time interval from t1 = 10.9 sto t2 = 22.4 s , the average acceleration of the dog has magnitude 0.49 m/s2 and direction 26.5 ∘measured from the +x−axis toward the +y−axis . A)At time t2 = 22.4 s , what is the y-component of the dog's velocity? B)What is the...
A dog running in an open field has components of velocity vx = 2.7 m/s and vy = -1.4 m/s at time t1 = 11.1 s . For the time interval from t1 = 11.1 s to t2 = 23.1 s , the average acceleration of the dog has magnitude 0.51 m/s2 and direction 31.0 ∘ measured from the +x−axis toward the +y−axis. 1) At time t2 = 23.1 s , what is the x-component of the dog's velocity? 2)...
A dog running in an open field has components of velocity vx = 2.7 m/s and vy = -1.2 m/s at time t1 = 10.5 s. For the time interval from t1 = 10.5 s to t2 = 21.6 s, the average acceleration of the dog has magnitude 0.54 m/s2 and direction 29.5 ∘ measured from the +x-axis toward the +y-axis. At time t2 = 21.6 s, what is the x-component of the dog's velocity? At time t2 = 21.6...
A dog running in an open field has components of velocity vx = 2.8 m/s and vy = -1.6 m/s at time t1 = 12.0 s . For the time interval from t1 = 12.0 s to t2 = 22.7 s , the average acceleration of the dog has magnitude 0.48 m/s2 and direction 32.0 ∘ measured from the +x−axis toward the +y−axis. A. At time t2 = 22.7 s , what is the x-component of the dog's velocity? B....
A dog running in an open field has components of velocity vx = 1.8 m/s and vy = -1.6 m/s at time t1 = 10.0 s . For the time interval from t1 = 10.0 s to t2 = 21.9 s , the average acceleration of the dog has magnitude 0.54 m/s2 and direction 30.0 ∘ measured from the +x−axis toward the +y−axis. a)At time t2 = 21.9 s, what is the x-component of the dog's velocity? b) At time...