Question

If velocity v of a particle moving on a straight line as a function of time t is given as v=5-t (m/s)then the distance covered by the particle in first 10 s is

Answer 1

Answer:

25 m

Explanation:

$\large \text{Given velocity of particle (v) = (5-t)m/sec and time (t) = 10sec}\\\\\\\large \text{we have to find distance here particle is in straight line so distance=displacement}\\\\\\\large \text{we know that change in velocity in time is equal to displacement }\\\\\\\large \text{v= \dfrac{ds}{dt} }\\\\\\\\\large \text{ds = v dt}\\\\\\\large ds=(5-t)dt$

$\large ds=(5-t)dt\\\\\\\large \text{take integration on both side }\\\\\\\ \large s=\int\limits^5_0 {(5-t)} \\\\\\\large v=(5-t)=0\\\\\\\large t=5sec\\\\\\\large \text{In first 5 sec velocity will be positive but 5 to 10 velocity will be negative}\\\\\\\large distance=x_s=x_{0-5}+x_{5-10}\\\\\\\large x_s=\int\limits^5_0 {(5-t)} \, dt \int\limits^{10}_5 {(t-5)} \, dt$

$\large x_s=[5t-\dfrac{t^2}{2}]^{5}_0 \,+[\dfrac{t^2}{2}-5t ]^{10}____5 \,  \\\\\\\large x_s=[25-\dfrac{25}{2} -0-0]+[\frac{100}{2}-50-\dfrac{25}{2}-25]\\\\\\\large x_s=\dfrac{25}{2} +\dfrac{25}{2} \\\\\\\large x_s=25 \ m\\\\\\\large \text{Thus total distance covered by particle in first 10 sec is 25 m.}$

Answer 2

Answer:

25 m

Explanation:

We are given that a particle moving on a straight line with velocity and velocity is a function of t is given by

$v=5-t$

We have to find the value of distance covered by the particle in 10 s

We know tha velocity =$\frac{ds}{dt}$

$ds=v\codtdt$

$ds=(5-t)dt$

When we substitute 5-t=0 then t=5 it means interval break at t=5 sec

The velocity from 0 to 5 sec is positive and from 5 to 10 sec then velocity will be negative

Now, $\int_{0}^{10}ds=\int_{0}^{10}\mid{5-t}\mid dt$ because we calculate distance and distance =speed multiply by time and speed is the absolute value of velocity

s=$\int_{0}^{5}(5-t)dt+\int_{5}^{10}(t-5)dt$

s=$[5t-\frac{t^2}{2}]^5_0+[-5t+\frac{t^2}{2}]^10_5$

s=$(25-\frac{25}{2})+(50-50+25-\frac{25}{2})$

s=50-25=25 m

Hence, distance covered by the particle =25 m