please answer questions 5,6, and 7. show all working, thank you D Question 5 10.
ID: 1878437 • Letter: P
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please answer questions 5,6, and 7. show all working, thank you
D Question 5 10.5 pts An object is initially traveling in the +x direction at 12 m/s. A constant acceleration of 9.6 m/s2 is applied in the -x direction. What is the t it travels, in meters, before it stops? If in the negative x direction, indicate so with a negative sign. DQuestion 6 10.5 pts An object is initially traveling in the tx direction at 12.5 m/s. A constant acceleration is applied to it and after it travels a displacement of 13.4 meters in the +x direction, it stops. What is the magnitude of the acceleration in m/s?? D Question 7 10.5 pts A car is initially moving at a velocity of 20.7 m/s in the +x direction and a constant acceleration is applied to it and it stops in 2.7 seconds. What is the acceleration in m/s?? If in the negative x direction, indicate so with a negative sign.Explanation / Answer
(5) Initial speed, u = 12 m/s
Acceleration is in the -x direction. So, a = -9.6 m/s^2
suppose distance travelled by the object before stop is 's'.
use the expression -
v^2 = u^2 + 2*a*s
put the values -
0 = 12^2 + 2*(-9.6)*s
=> s = 12^2 / (2*9.6) = 7.5 m (Answer)
(6) Here, u = 12.5 m/s, s = 13.4 m, accleration a = ?
use the expression -
v^2 = u^2 + 2*a*s
put the values -
0 = (12.5)^2 + 2*a*13.4
=> a = - (12.5)^2 / (2*13.4) = -5.83 m/s^2
Negative sign shows that the acceleration is in the negative x direction.
(9) Here, initial velocity, u = 20.7 m/s, t = 2.7 s, a = ?
use the expression -
v = u + a*t
0 = 20.7 + a*2.7
=> a = -20.7 / 2.7 = -7.67 m/s^2
Negative sign shows that the acceleration is in the negative x direction.
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