Question

the displacement of x of a particle moving in One dimension under the action of constant force is related to the time by the equation X =(5t ^2 + 3t-5 ) m where x is in metres and T is in seconds. find the velocity of the particle at 1(t) = 3 s 2(t)=6s ​

Answer 1

Answer:

1 ) 33 m/s

2 ) 63 m /s

Step-by-step explanation:

$x = 5 {t}^{2} + 3t - 5 \\ \\ \frac{dx}{dt} = 10t + 3 \\ \\ a) \: \: \: at \: t = 3 \\ v = 33 \: \: \: m {s}^{ - 1} \\ \\ \\ b) \: \: \: \: \: at \: \: t = 6 \\ v = 63 \: \: \: m {s}^{ - 1}$