A car is travelling in the speed of 12m/s increases its speed to 22m/s in the 1st 10 seconds . it decreases its speed to 18m/s in next 6 seconds. find the acceleration of the car in both cases
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So first car is accelerating at a speed of 12 m/s which changes to 22 m/s in 10 seconds, acceleration is given by formula a = dv/dt which further equates to a = (22-12)/10 => 1 m/s^2
In second case acceleration a is given by a= (18-22)/6 => -1 m/s^2 (i.e car is deaccelerating/retarding)
In second case acceleration a is given by a= (18-22)/6 => -1 m/s^2 (i.e car is deaccelerating/retarding)
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Answer:
First Case : 1 m/s²
Second Case : 0.66 m/s²
Step-by-step explanation:
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