Question

an object moving with a velocity of 5m/s comes to rest in 10 seconds ,find the acceleration and distance covered of class​

Answer 1

Given :

Initial velocity : 5 m/s

Final velocity : 0 m/s

Time taken : 10 s

To find :

The retardation and the distance travelled.

Solution :

Accerlation is the rate of change of velocity.

• It is a vector quantity having both magnitude as well as direction.
• SI unit of acceleration is metre per second square.

As we are provided with initial velocity, final velocity and time taken we can use 1st equation of motion that is,

v = u + at

where, v denotes final velocity, u denotes initial velocity, a denotes acceleration and t denotes time taken.

By substituting all the given values in the equation,

➙ v = u + at

➙ 0 = 5 + a(10)

➙ -5 = 10a

➙ a = -0.5 m/s²

If the speed is decreasing with time then accerlation is negative. The negative accerlation is called deceleration or retardation.

Thus, the acceleration of the bus is - 0.5 m/s².

Now, using 2nd equation of motion that is,

s = ut + 1/2at²

By substituting the values in the equation,

➙ s = ut + 1/2at²

➙ s = (5)(10) + 1/2(-0.5)(10)²

➙ s = 50 - 25

➙ s = 25 m

Thus, the distance travelled by the bus is 25m.