Question

A hockey player strikes a puck, giving it an initial velocity of 14.0 m/s in the positive x-direction. The puck slows uniformly to 6.50 m/s when it has traveled 45.0 m. (Take the direction of initial motion as the positive direction. Indicate the direction with the sign of your answer.)

Answer 1

a) $-1.71 m/s^2$

b) 7.7 m/s

c) 4.39 s

Explanation:

a)

The acceleration of an object is the rate of change of velocity of an object.

In this problem, the acceleration of the puck can be found using the following suvat equation:

$v^2-u^2=2as$

where:

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled

For the puck in this problem:

u = 14.0 m/s

v = 6.50 m/s

s = 45.0 m

So, the acceleration is:

$a=\frac{v^2-u^2}{2s}=\frac{6.50^2-14.0^2}{2(45.0)}=-1.71 m/s^2$

b)

The velocity of the puck at time t can be found by using another suvat equation:

$v=u+at$

where

u is the initial velocity

a is the acceleration

t is the time elapsed

v is the final velocity

Here, we have:

u = 14.0 m/s

$a=-1.71 m/s^2$ (found in part a)

Therefore, the velocity of the puck after t = 3.70 s is:

$v=14.0+(-1.71)(3.70)=7.7 m/s$

c)

Here we want to find the time taken for the puck to travel a distance of

s = 45.0 m

To solve this part, we can use again the suvat equation:

$v=u+at$

Where in this case, we use:

u = 14.0 m/s is the initial velocity

v = 6.50 m/s is the final velocity when the puck has travelled 45.0 m (this information is given in the question)

$a=-1.71 m/s^2$ is the acceleration (found in part a)

Therefore, by re-arranging the equation, we find the time taken to cover 45.0 m:

$t=\frac{v-u}{a}=\frac{6.50-14.0}{-1.71}=4.39 s$