At a certain time a particle had a speed of 18m/s in thepositive direction, and

Question

At a certain time a particle had a speed of 18m/s in thepositive direction, and 2.4 seconds later its speed was 30 m/s inthe opposite direction. (a) What is the average acceleration of the particleduring this 2.4 second interval? (b) What is the total distance traveled during the same2.4 second time interval? At a certain time a particle had a speed of 18m/s in thepositive direction, and 2.4 seconds later its speed was 30 m/s inthe opposite direction. (a) What is the average acceleration of the particleduring this 2.4 second interval? (b) What is the total distance traveled during the same2.4 second time interval?

Explanation / Answer

.    Initial speed(u) = 18 m /s .    Final Speed (v) = - 30 m /s ( since it is inopposite direction) .    Time (t) = 2.4 s . (a) Average acceleration is : a = ( v - u ) / t = ( - 30 - 18 ) /2.4 = -20 m /s2   . (b) Distance traveled is : .    s = ut + (1/2) at2    .    = ( 18 * 2.4 ) + ( 0.5 * (-20) *(2.4)2   .    = ---------- m . Solve the above. . Hope this helps u! .    Final Speed (v) = - 30 m /s ( since it is inopposite direction) .    Time (t) = 2.4 s . (a) Average acceleration is : a = ( v - u ) / t = ( - 30 - 18 ) /2.4 = -20 m /s2   . (b) Distance traveled is : .    s = ut + (1/2) at2    .    = ( 18 * 2.4 ) + ( 0.5 * (-20) *(2.4)2   .    = ---------- m . Solve the above. . Hope this helps u!

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