Question

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

Answer 1

The value is
`t_1 = 9 \ s`

Step-by-step explanation:

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

`v = u + at`

Here u = 0 \ m/s

a = 13.5
`m/s^2`

So

`v = 0 + 13.5 * 3.10`

=>
`v = 41.85 \ m/s`

The is distance it covers at this time is

`s = u * t + (1)/(2) a * t^2`

=>
`s = + (1)/(2) * 13.5 * 3.10^2`

=>
`s =64.87`

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

So

`v_f = v + a_1t_1`

Here
`a_1 =  - 4.65`, the negative sign shows that it is deceleration

So

`0  =  41.85  - 4.65 *  t_1`

=>
`t_1 = 9 \ s`