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

Answer:

The value is $t_1 = 9 \ s$

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

$s = u * t + \frac{1}{2} a * t^2$

=> $s = + \frac{1}{2} * 13.5 * 3.10^2$

=> $s =64.87$

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

           $0 = 41.85 - 4.65 * t_1$

=> $t_1 = 9 \ s$