Question

A block sliding on a horizontal surface has an initial speed of 0.5 m/s. The block travels a distance of 1 m as it slows to a stop. What distance would the block have traveled if its initial speed had been 1 m/s

Answer 1

Answer:

4m (with the assumption that the deceleration is same in both cases)

Explanation:

The third equation of a uniformly decelerated motion is used in solving this problem.

$v^2=u^2-2as.................(1)$

where v is the final velocity, u the initial velocity, s the distance covered and a the deceleration.

The negative sign is due to the fact that the body under consideration is decelerating.

Given;

u = 0.5m/s

v = 0m/s (because it comes to rest)

s = 1m

a = ?

Substituting all values into equation (1);

$0^2=0.5^2-2*a*1\\2a=0.5^2\\2a=0.25\\a=\frac{0.25}{2}\\a=0.125m/s^2$

Assuming the block maintained this same deceleration, but with an initial velocity of 1m/s, then the distance travelled would be as follows;

u = 1m/s

v  = 0m/s

s = ?

a = $0.125m/s^2$

By still using equation (1), we obtain the following;

$0^2=1^2-2*0.125*s\\0.25s=1\\s=\frac{1}{0.25}\\s=4m$

Answer 2

Explanation:

Below is an attachment containing the solution.