Engineering Find the distance given a function for v(t)

[tremain74]

Homework Statement

Distance by speed of a particle

Relevant Equations

I want to see if I am using the right technique for this problem. The speed of a particle is given by v = 2*t + 5*t^2, where t is in seconds. What distance does it travel while its speed increases from 7 to 99 m/s?

My solution: v = 2*t + 5*t^2. dx/dt = 2*t + 5*t^2. Therefore dx = (2t + 5t^2)dt. After taking the anti derivative my equation is s = t^2/2 + 5/3(t^3).

 

[FactChecker]

That's good so far. Your problem says that the speed goes from 7 to 99, but your equation for $s$ is in terms of time, $t$. So what is your next step?

PS. A little knit-picking: 5/3(t^3) is easily confused as $\frac {5}{3(t^3)}$. It would be better to write 5(t^3)/3.
Also, an anti derivative includes a constant, $c$. The constant does not change the value of the definite integral.

 

[tremain74]

At first I was going to use the values 7m /s to 99 m/s as the numbers to plug into the equations to solve for the distance but these values are for velocity and not for seconds.

 

[Frabjous]

You ignored the 2 in the 2t term.

You can get times from the velocity equation using the given velocities.