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jenyasd209 [6]
3 years ago
14

A test car starts from rest on a horizontal circular track of 115-m radius and increases its speed at a uniform rate to reach 90

km/h in 17 seconds. Determine the magnitude a of the total acceleration of the car 13 seconds after the start
Physics
1 answer:
Wewaii [24]3 years ago
3 0

Answer:

a= 3.49 m/s^2

Explanation:

magnitude of total acceleration = sqrt{radial acceleration^2+tangential acceleration^2}.

we know that tangential acceleration a_t= change in velocity /time taken

now 90 km/h = 25 m/s

a_t = 25/17 = 1.47 m/s^2.

radial acceleration a_r = v^2/r

v= a_t×t = 1.47×13 = 19.11 m/s

a_r = 19.11^2/115=  3.175

now,

a= \sqrt{a_t^2+a_r^2}

a= \sqrt{1.47^2+3.175^2}

a= 3.49 m/s^2

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