Answer to Question #101624 in Physics for kwikiriza steven

Question

A car moves from rest accelarating at 4m/s within 4 sec during the next 10 sec the car moves uniformly after what they break and the car slow down at 8m/s until it stops. find the total distance moved by the car

Answer 1

1.

$$d_1=0.5a_1t_1^2=0.5(4)(4)^2=32\ m$$

$$v=a_1t_1=(4)(4)=16\frac{m}{s}$$

2.

$$d_2=vt_2=(16)(10)=160\ m$$

3.

$$v^2-0^2=2a_2d_3$$

$$16^2=2(8)d_3$$

$$d_3=16\ m$$

So, the total distance moved by the car:

$$d=32+160+16=208\ m$$