Question

A particle travels along a straight line, in a given direction, with constant acceleration. At instant t0 = 0, the magnitude of its velocity is v0 = 5 m/s; at time t = 10s, v = 25m/s. determine:
a) The type of motion of the particle
b) Acceleration
c) The function of velocity in relation to time
d) The velocity at time t = 8.0 s
e) The instant of time in which the velocity module is v = 15m/s

Answer 1

a) The motion described by the particle is undoubtedly a uniformly accelerated rectilinear motion, since the trajectory is rectilinear and the acceleration is constant.

b) In order to calculate the acceleration, we must apply the formula mentioned above, in such a way that the acceleration gives us:

$\boldsymbol{a=\dfrac{v-v_{o}}{t-t_{o} }=\dfrac{25-5}{10-0 }=\dfrac{20}{10}=2 \ m/s^2 }$

Our acceleration is 2 meters per second squared.

c) We are asked to calculate the function of velocity in relation to time, we simply substitute in the formula.

$\boldsymbol{v=v0+at }\\\boldsymbol{v=2+5t }$

Very easy!!.

d) To know what speed the particle will have at the instant of t = 8s, it is enough to substitute the value of “t” in the previous formula.

$\boldsymbol{v=5+2(8)=5+16=21 \ m/s }$

So the speed at time t = 8s is 21 m/s²

e) In this case we are asked to determine at what instant of time the particle will have a speed of 15 m/s, we replace this value in the formula, simply clearing the variable "t", that is:

$\boldsymbol{v=v_{0}+at }$

Clearing "t"

$\boldsymbol{t=\dfrac{v-v_{o}}{a} }$

Substituting the speed value

$\boldsymbol{t=\dfrac{v-v_{o}}{a}=\dfrac{15-5}{2}=\dfrac{10}{2}=5 \ seg }$

That is to say that when the particle has a speed of 15 m/s, it will happen exactly at 5 seconds.