Question

An object having a velocity 40m/sec2 is accelerated at a rate of 1.2m/s2 for 5.0sec. Find the distance traveled during the time of acceleration

Answer 1

ANSWER:

initial velocity (u) = 4ms^-1

acceleration (a) = 1.2ms^-2

time (t) = 5s

SOLUTION

we know a kinematical equation

s = ut +1/2at ^2

= 4 (5) +1/2 (1.2)(5)^2

= 20 + (0.6)(25)

= 20 + 15

= 35m

there distance = 35m

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