Physics kinematics help
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Physics kinematics help

[From: ] [author: ] [Date: 12-06-07] [Hit: ]
5.33 m/s^2-u=20m/s, T=4,a=16/6 = 2.......
I can't figure out how to solve this for the life of me. Please show me the process. Thank you

At time t = 0.00 s, a car is traveling along a straight line at a speed of 20 m/s with constant acceleration. The car travels 56.0 m during the time interval between t = 2.00 s and t = 4.00 s. What is the acceleration of the car?

A. 0.00 m/s^2
B. 2.67 m/s^2
C. 14.0 m/s^2
D. 7.00 m/s^2
E. 5.33 m/s^2

-
u=20m/s, T=4, t=2
let S4 is the distance travelled by the car in 4s & S2 distance travelled in 2s
S4-S2=56
uT+1/2aT^2 -(ut+1/2at^2) = 56
80 + 8a - 40 - 2a=56
40 +6a=56
a=16/6
= 2.6m/s^2
hence (b) is correct answer
1
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