If an object with uniform acceleration (or deceleration) has a speed of #3 m/s# at #t=0# and moves a total of 58 m by #t=6#, what was the object's rate of acceleration?

Answer 1

a=20/9 m/s^2
a=-20/9 m/s^2

#58=(3+k)/2*6"; "58=(3+k)*3"; "58=9+3*k#
#49=3*k"; "k=49/3#
#"rate of acceleration is equal tangent of graph"#
#a=(3-k)/6" or " a=(k-3)/a#
#a=(3-49/3)/6" ; " a=-20/9 m/s^2#
#a=(49/3-3)/6" ; " a=20/9 m/s^2#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Parker Abbott

To find the acceleration, you can use the equation of motion:

[v = u + at]

where: (v) = final velocity = 3 m/s (u) = initial velocity = 0 m/s (since the object starts from rest) (a) = acceleration (which we need to find) (t) = time = 6 seconds

Rearranging the equation to solve for acceleration:

[a = \frac{v - u}{t}]

[a = \frac{3 - 0}{6}]

[a = 0.5 , \text{m/s}^2]

So, the object's acceleration is (0.5 , \text{m/s}^2).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.