Question

a particle is moving on a straight line with speed v= 30t^2 +4t +2.the average speed from t=0 to t=5s.​

Answer 1

Answer:262m/s

Explanation-on integration of 30t^2+4t+2 with upper limits and lower limits as 5and 0respectively we get answer 1310

Avg.speed-total distance /total time taken

So ,1310÷5=262m/s

Answer 2

Answer:262 m/s

Explanation:

Given

$v=30t^2+4t+2$

Average speed from t=0 to t=5 s

Average speed is given by

$v_{avg}==\frac{\int_{0}^{5}(30t^2+4t+2)dt}{\int_{0}^{5}dt}$

$v_{avg}==\frac{10\left ( t^3\right )_0^5+2\left ( t^2\right )_0^5+2\left ( t\right )_0^5}{\left ( t\right )_0^5}$

$v_{avg}=\frac{1310}{5}=262 m/s$