Question

A block slides to a stop as it goes 47 m across a level floor in a time of 6.35 s. a) What was the initial velocity? b) What is the acceleration?

Answer 1

(a) The initial velocity is, 14.8 m/s

(b) The acceleration is,
`-2.33m/s^2`

Explanation :

By the 1st equation of motion,

`v=u+at` ...........(1)

where,

v = final velocity = 0 s

u = initial velocity

t = time = 6.35 s

a = acceleration

The equation 1 will be:

`0=u+a(6.35}`

`u=-6.35a` ..........(2)

By the 2nd equation of motion,

`s=ut+(1)/(2)at^2` ...........(3)

where,

s = distance = 47 m

Now substitute equation 2 in 3, we get:

`47=(-6.35a)* (6.35)+(1)/(2)* a* (6.35)^2`

By solving the term, we get:

`a=-2.33m/s^2`

The acceleration is,
`-2.33m/s^2`

Now we have to calculate the initial velocity.

Using equation 2, we gte:

`u=-6.35a`

`u=-6.35s* (-2.33m/s^2)`

`u=14.8m/s`

The initial velocity is, 14.8 m/s