Question

a body moves with uniform velocity of u = 7m/s from t =0 to t = 1.5 sec. For t> 1.5sec, it starts moving with a acceleration of 10m/s2. The distance travelled between t = 0 to t =3 sec. will be ?​

Answer 1

Answer:

32.25 m

Explanation:

For t = 0 to t=1.5,

distance= 7*1.5= 10.5m

For t=1.5 to t=3 sec

distance= 7*1.5 + 1/2* 10* 1.5*1.5 = 21.75m

So, total distance= 21.75+ 10.50= 32.25m

Answer 2

Given that :

Initial velocity of the body, u = 7 m/s from = 0 to t = 1.5 s

It starts moving with a acceleration of 10 m/s².

To find,

The distance travelled between t = 0 to t =3 sec.

Solution,

The distance traveled by the body from t = 0 to t = 1.5 s will be :

$d=ut\\\\d=7\times 1.5\\\\d=10.5\ m$

Now the distance covered in next 1.5 seconds is given by :

$d=ut+\dfrac{1}{2}at^2\\\\d=7\times 1.5+\dfrac{1}{2}\times 10\times (1.5)^2\\\\d=21.75\ m$

Total distance traveled between 0 to 3 seconds is :

$d=10.5+21.75\\\\d=32.25\ m$

So, the total distance traveled between 0 to 3 seconds 32.25 meters.

Learn more,

Kinematics

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