Question

an object accelerates from rest to velocity 30 m/s in 10 s , then find distance covered by object during next 10 s ​

Answer 1

Answer:

450m

Explanation:

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Answer 2

Answer:

The distance covered by object during next 10s ​is 450m.

Explanation:

Given initial velocity, $u=0$

          final velocity, $v=30m/s$

          time taken, $t=10 s$  

Using equation of motion,

                      $v=u+at\implies30=0+a\times 10$

                                         $\implies a=\frac{30}{10} =3m/s^{2}$

Hence, the acceleration of the object, is $3m/s^{2}$

For next 10sec,  initial velocity, $u=30m/s$

Using equation of motion,

                      $s=ut+\frac{1}{2} at^{2} \implies s=30\times 10+\frac{1}{2}\times 3\times (10)^{2}$

                                                        $=300+150=450m$

Therefore, the distance covered by object during next 10s ​is 450m.