Question

A dog initially running at a speed of 2 m/s accelerates at constant rate of 1.5 m/s2 for 5 s.Calculate:(1) distance covered by it from start (2) the final velocity of the dog?

Answer 1

(1) Total distance covered is `28.75m` and (2) the final velocity is `9.5m//s`

Explanation:

We know the formula,`v=u+at` where `v` is final velocity, `u` is initial velocity, `a` is acceleration & `t` is time.
Here we get, `v=?,u=2ms^-1,a=1.5ms^-2,t=5s`
Hence,(2) Final velocity of the dog is `(2+1.5xx5)=9.5ms^-1`
And the another formula to determine the covered distance is
`v^2=u^2+2as` where `s` is displacement.
Hence,here displacement is
`(v^2-u^2)/(2a)=(9.5^2-2^2)/(2xx1.5)=28.75m`
Hope it helps......
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