Answer
The speed of the car after $1.3s$ is $6.92m/s$.
Work Step by Step
1) We have initial velocity $v_0=16.1m/s$ and the elapsed time $t=1.3s$. To calculate the speed $v$ $1.3s$ after braking, we use the formula
$$v=v_0+at$$
The missing component here is deceleration $a$.
2) Find $a$
When the driver applies the brake, there is no further force applied on the car by the engine. Therefore, kinetic frictional force takes over completely and brings the car to a halt.
$$f_k=\mu_kF_N=\mu_km_{car}g=0.72\times m_{car}\times9.8=7.06m_{car}$$
Also, according to Newton's 2nd Law: $$f_k=m_{car}\times a$$ with $a$ being the deceleration of the car after braking
Combining 2 equations, we have $$am_{car}=7.06m_{car}$$ $$a=7.06m/s^2$$
But since $a$ is the deceleration, $a=-7.06m/s^2$
3) Calculate the speed of the car after $1.3s$ $$v=16.1-7.06\times1.3=6.92m/s$$
