Question

In the graph shown below, t₀ = 2.1 seconds, the object stops at 8 seconds, and t' = 22 seconds.  First, find the value of the object's velocity at 2.1 seconds such that the object's total displacement from 0to 22 seconds is -55.7 meters.  Then, using this value find the acceleration from t = 0 to 2.1 seconds.  Answer in m/s².

please show work and explain

Answer 1

Here ,

t0 = 2.1 s

t1 = 8 s

t' = 2.1 s

let the velocity of object at t = 2.1 s is v0

Now, for the velocity at t = 22 s

acceleration during t = 2.1 s t 22 s is a

a = -v/(8 - 2.1)

a = -0.169 * v0

Now, velocity at t = 22 s

v = - (22 - 8) * 0.169 * v0

for the total displacement

0.5 * 2.1 * v0 + 0.5 * v0 * (8 - 2.1) - 14 * 0.169 * v0 * 0.5 * 14 = -55.7

solving for v0

v0 = 4.43 m/s

Now, for the acceleration

acceleration during first 2.1 s = v0/2.1

acceleration during first 2.1 s = 4.43/2.1

acceleration during first 2.1 s = 2.11 m/s^2

the acceleration during first 2.1 s is 2.11 m/s^2