Question

The velocity of a particle is given by v=20t² - 100t + 50, where v is in meters per second and t is in seconds. Evaluate the velocity when a is zero and the distance travelled at that instant.

Answer 1

Answer:

velocity =  50m/s

distance = -83.33m

Explanation:

The velocity of the particle is given by;

v = 20t² - 100t + 50        -------------------(i)

Since acceleration is the time rate of change in velocity, to get the acceleration (a), we find the derivative of equation (i) with respect to t as follows;

a = $\frac{dv}{dt}$ = $\frac{d(20t^{2} - 100t + 50)}{dt}$

a = 40t - 100            ------------------(ii)

Now, when a = 0, let's find the time t by substituting the value of a into equation (ii) as follows;

0 = 40t - 100

=> 40t = 100

=> t = $\frac{100}{40}$

=> t = 2.5seconds.

This means that at t = 2.5 seconds, the acceleration of the particle is zero(0)

(a) Now, to get the velocity at this instant (t = 2.5s), substitute the value of t into equation (i) as follows;

v = 20(0)² - 100(0) + 50

v = 0 - 0 + 50

v = 50 m/s

Therefore, the velocity when a is zero is 50m/s

(b) To get the distance (s) travelled at that instant, we integrate equation (i) as follows;

s = $\int\limits^a_b {v} \, dt$      -----------------------(iii)

Where;

a = the time instant = 2.5 seconds

b = the initial time instant = 0

v =  20t² - 100t + 50

Substitute these values into equation (iii) as follows;

s = $\int\limits^a_b {(20t^{2} -100t + 50)} \, dt$

s = $\frac{20t^{3} }{3}$ - $\frac{100t^{2} }{2}$ + 50t $|^{a}_{b}$

Substitute the values of a and b as follows;

s = [$\frac{20(2.5)^{3} }{3}$ - $\frac{100(2.5)^{2} }{2}$ + 50(2.5)]  - [$\frac{20(0)^{3} }{3}$ - $\frac{100(0)^{2} }{2}$ + 50(0)]

s =  [$\frac{20(2.5)^{3} }{3}$ - $\frac{100(2.5)^{2} }{2}$ + 50(2.5)] - 0

s = 104.17 - 312.5 + 125

s = -83.33m

Therefore, the distance traveled at that instant is -83.33m

Answer 2

Explanation:

It is given that,

The velocity of a particle is given by :

$v=20t^2-100t+50$

Where

v is in m/s and t is in seconds

Let a is the acceleration of the object at time t. So,

$a=\dfrac{dv}{dt}$

$a=\dfrac{d(20t^2-100t+50)}{dt}$

$a=40t-100$

When a = 0

$40t-100=0$

t = 2.5 s

a is zero at t = 2.5 s. Velocity, $v=20(2.5)^2-100(2.5)+50$

v = -75 m/s

Since, $v=\dfrac{ds}{dt}$, s is the distance travelled

$s=\int\limits{vdt}$

$s=\int\limits{(20t^2-100t+50)dt}$

$s=\dfrac{20t^3}{3}-50t^2+50t$

At t = 2.5 s, $s=\dfrac{20(2.5)^3}{3}-50(2.5)^2+50(2.5)$

s = −83.34 m

Hence, this is the required solution.