Question

A bear is walking through a forest with a velocity of 1.5 m/s when it spots some honey 200 meters ahead. Three seconds later, it is running toward the honey with a velocity of 7.5 m/s. What is the bear's acceleration?

Answer 1

Using SUVAT

u = 1.5 m/s
v = 7.5 m/s
a = ?
t = 3 s

v = u + at
(v - u)/t = a

insert the numbers that correspond to the letters into that last equation to get the acceleration (a)

Answer 2

The bear's acceleration is
`2(m)/(s^(2))`

Step-by-step explanation:

The magnitude of the acceleration can be calculated using the following equation :

`a=(dv)/(dt)` (I)

Where ''a'' is the acceleration

Where ''dv'' is the speed variation

Where ''dt'' is the time variation

In this exercise, the time variation is equal to 3 seconds because it is the amount of time in which the bear accelerated from
`1.5(m)/(s)` to
`7.5(m)/(s)`

The speed variation is equal to :

`dv=vf-vi`

Where ''vf'' is the final speed and ''vi'' is the initial speed.

Finally, we can calculate the bear's acceleration using the equation (I) :

`a=(dv)/(dt)`

`a=(7.5(m)/(s)-1.5(m)/(s))/(3s)`

`a=(6(m)/(s))/(3s)`

`a=2(m)/(s^(2))`

We find that the bear's acceleration is
`2(m)/(s^(2))`