The table below contains data for 1-D motion described by: xo-3 m, vo-5 m/s, a -

Question

The table below contains data for 1-D motion described by: xo-3 m, vo-5 m/s, a - 2 m/s2. The table has the same time interval between all rows. A few table cells are filled in with correct numbers for reference. Others cells are left blank. Some cells contain letters which are referenced in the questions which follow. Choose the correct answer in each case. It (sec) v(m/s)x(m) l 0.00 5.003.00I 0.01 AI 004 5.08 3.20 I ICI D E I A] (m/s) I Choose ] time interval between rows () [Choose ] B) (m) l Choose I Choose] DI(n/s) T Choose j Choost

Explanation / Answer

From the data given , we can write the following

Initial time t0 = 0

Initial speed v0 = 5 m/s

Acceleration is uniform and equal to a = 2 m/s2

Initial position is x0 = 3 m

We can easily identify from the data and values given in the table that time interval t = 0.01 s

Now first we try to calculate all the values possible in the table and then write down the required answers.

Let us denote the time at a particular instant with t so that

t0 = 0 s , t1 = 0.01 s , t2 = 0.02 s , t3 = 0.03 s ,

t4 = 0.04 s , t5 = 0.05 s

Let us calculate the speeds at various instances using the first kinematics equation , v = v0 + a t

At a time t1 = 0.01 s ,

Speed v1 = v0 + a t1 = 5 + 2 × 0.01 = 5.02 m/s2

At a time t2 = 0.02 s ,

Speed v2 = v0 + a t2 = 5 + 2 × 0.02 = 5.04 m/s

At a time t3 = 0.03 s ,

Speed v3 = v0 + a t3 = 5 + 2 × 0.03 = 5.06 m/s

At a time t4 = 0.04 s ,

Speed v4 = v0 + a t4 = 5 + 2 × 0.04 = 5.08 m/s

At a time t5 = 0.05 s ,

Speed v5 = v0 + a t5 = 5 + 2 × 0.05 = 5.10 m/s

Let us calculate distances at various instances using the second kinematics relation

x = x0 + v0t + 1/2 a t2

At a time t1 = 0.01 s ,

Distance x1 = x0 + v0 t1 + ( 1/2 ) × a t12

= 3 + ( 5 × 0.01 ) + ( 1/2 × 2 × 0.012 )

= 3.0501 m

= 3.05 m

Similarly distance x2 = x0 + v0 t2 + 1/2 a t22

= 3 + ( 5 × 0.02 ) + ( 1/2 × 2 × 0.022 )

= 3.1004 m

= 3.10 m

Distance x3 = x0 + v0 t3 + 1/2 a t32

= 3 + ( 5 × 0.03 ) + ( 1/2 × 2 × 0.032 )

= 3.1509 m

= 3.15 m

Distance x4 = x0 + v0 t4 + 1/2 a t42

= 3 + ( 5 × 0.04 ) + ( 1/2 × 2 × 0.042 )

= 3.2016 m

= 3.20 m

Distance x5 = x0 + v0 t5 + 1/2 a t52

= 3 + ( 5 × 0.05 ) + ( 1/2 × 2 × 0.052 )

= 3.2525 m

= 3.25 m

Now let us find the required answers

1. A is the speed at time t1 , so v1 = 5.02 m/s is the answer for A.

2. Time interval is t = 0.01 s

3. B is the distance traveled in a time t2 = 0.02 s , so answer for B is x2 = 3.1004 m = 3.10 m

4. C is the time t5 = 0.05 s

5. D is the speed at time t5 , which is v5 =5.10 m/s

6. E is the corresponding distance x5 = 3.25 m

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