Question

A body starts from rest and with a uniform
acceleration of 10 ms-2 for 5 seconds. During the
next 10 seconds it moves with uniform velocity. The
total distance travelled by the body is :-​

Answer 1

Answer:-

S = 625 m

Given :-

u = 0 m/s

t = 5s

a = 10 m/s²

To find :-

The total distance travelled.

Solution:-

First,

we have to find the acceleration produced in the body.

From first eq. of motion:-

→$v = u + at$

→$v = 0 + 10 \times 5$

→$v = 50 m/s$

Distance travelled by body :-

→$2as = v^2 - u^2$

→$2 \times 10 \times s = (50) ^2 - (0) ^2$

→$20s = 2500$

→$s = \dfrac{2500}{20}$

→$s = 125 m$

Now, The body moves with the constant velocity.

by using distance formula,

→$\mathsf{s" = vt }$

→$\mathsf{s" = 50 \times 10}$

→$\mathsf{ s" = 500 m }$

Total distance travelled by body :-

→$\mathsf{S = s + s"}$

→$\mathsf{S = 125 + 500}$

→$\mathsf{S = 625 m }$

hence,

Distance travelled by body is 625m.

Answer 2

$\mathfrak{\large{\underline{\underline{Answer:-}}}}$

The total distance travelled is  =  625m.

$\mathfrak{\large{\underline{\underline{Explanation:-}}}}$

To Find:

The total distance travelled by the body.

Given that,

The initial velocity (u) = 0 m/s

The time is (t) = 5s

The acceleration (a) = 10 m/s²

Solution :

We know that,

Equation of motion :

★ $\boxed{\sf {v=u+at}}$

So,

Velocity is :

$\sf \longrightarrow v=0+10(5)$

$\sf \longrightarrow 50 ms^-^1$

The distance travelled in the first 5s is:

$\longrightarrow \sf 2as = v^2 - u^2\\\longrightarrow \sf 0\times5+12\times10\times52\\\longrightarrow \sf 125m$

The distance travelled in the 10s is :

By Distance formula :

$\longrightarrow \sf s^1=vt_1\\\longrightarrow \sf 50\times 10\\\longrightarrow \sf 500m$

Total Distance :

$\longrightarrow \sf d=s+s_1\\\longrightarrow \sf 125+500\\\longrightarrow \sf 625m.$

Hence,

The total distance travelled by body is 625m.