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

Answer :

(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,

$s=ut+\frac{1}{2}at^2$     ...........(3)

where,

s = distance = 47 m

Now substitute equation 2 in 3, we get:

$47=(-6.35a)\times (6.35)+\frac{1}{2}\times a\times (6.35)^2$

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:

$u=-6.35a$

$u=-6.35s\times (-2.33m/s^2)$

$u=14.8m/s$

The initial velocity is, 14.8 m/s