How to find the constant acceleration of this question!?

Question

An object travelling with constant acceleration has a velocity of 17cm/s at t=2s and is 95 cm from its initial position at t=5s. What is the constant acceleration?

Answer 1

Let a= acceleration (a constant). Since acceleration is the derivative of velocity, by integrating the acceleration, we can get velocity.

Int(a dt) = at + C ; a is a constant with t as the variable

In other words, velocity (v) is equal to a*t +C. At t=2, v(t) is 17 so this means that a*2 + C = 17 (or 17 = 2a +C)

If we write C is terms of a we get C = 17 - 2a.

Let position be x(t). You know that if at t=5, it is 95cm from it's original position, then x(5) - x(0) = 95. In terms of integration, this is the definite integral of velocity from 0 to 5. We do that next (integrating velocity from 0 to 5) :

Int( at + C dt) from 0 to 5 is [ (at^2)/2 + Ct] from 0 to 5. Remember that C= 17 -2a. Substitute that:

[(at^2)/2 + (17 -2a)t] from 0 to 5.

Next we use the fundamental theorem of calculus to evaluate this:

= [(a *5^2)/2 + (17-2a)*5] - [(a*0^2)/2 +(17-2a)*0]

= [25a/2 + 85 -10a] - [0] = 2.5a + 85.

This is equal to the change in position (95)

2.5a + 85 = 95

Solving for a yields a = 4. The unite for acceleration in this problem is cm/ second^2

So acceleration is 4 cm/ (sec)^2

Answer 2

Let's say the object had speed v0 at t = 0 seconds.



Then its average velocity over the 1st 2 seconds is (17 + v0)/2, and it travels v * t = 2 * (17 + v0)/2 cm in those two seconds.



So at t = 2, it is 17 + v0 cm from the starting position. Its acceleration is a = (17-v0)/2



The distance it travels in 5 seconds is v0*t + (1/2)at^2 = v0*5 + (1/2)*(17-v0)/2 * 5^2 = 95



You can solve this equation for v0, and use that to get the acceleration.



Edit: No, actually it does make sense once you write it all down. I got an answer. That's a linear equation which works out to a whole-number value for v0 and for a.

Answer 3

17 = u + 2a ...and 95 = 5u + 12.5 a ....solving for u and a yields



a =4 cm/sec^2....and u = 9 cm/sec



constant acceleration is 4 cm/sec^2





used laws of motion.

Answer 4

Those two sets of data don't seam to be consistent with each other, the Velocity would yield 8.5cm/sec2 and the displacement would yield 7.6cm/sec2. Something doesn't jive.

Answer 5

acceleration



38/5