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Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a large toy rocket to the back of a sled and take the modified sled to a large, flat, snowy field. You ignite the rocket and observe that the sled accelerates from rest in the forward direction at a rate of 13.513.5 m/s2 for a time period of 3.103.10 s. After this time period, the rocket engine abruptly shuts off, and the sled subsequently undergoes a constant backward acceleration due to friction of 4.654.65 m/s2. After the rocket turns off, how much time does it take for the sled to come to a stop

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okpalawalter8

Answer:

The value is [tex]t_1 = 9 \ s [/tex]

Explanation:

Generally the velocity attained by the sled after t = 3.10 s is mathematically evaluated using the kinematic equation as follows

[tex]v = u + at[/tex]

Here u = 0 \ m/s

a = 13.5 [tex]m/s^2[/tex]

So

[tex]v = 0 + 13.5 * 3.10 [/tex]

=> [tex]v = 41.85 \ m/s [/tex]

The is distance it covers at this time is

[tex]s = u * t + \frac{1}{2} a * t^2[/tex]

=> [tex]s = + \frac{1}{2} * 13.5 * 3.10^2[/tex]

=> [tex]s =64.87 [/tex]

Now when sled stops its the final velocity is [tex] v_f = 0 m/s [/tex] while the initial velocity will be the velocity after its acceleration i.e [tex]v = 41.85 \ m/s [/tex]

So

[tex]v_f = v + a_1t_1[/tex]

Here  [tex]a_1 =  - 4.65[/tex], the negative sign shows that it is deceleration

So

           [tex]0  =  41.85  - 4.65 *  t_1[/tex]

=> [tex]t_1 = 9 \ s [/tex]

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