Question #98122

a car takes off from a stop sign with zero initial velocity but with nonzero acceleration of 6.50 m/s2.
a. what is the car's speed after 5.00 seconds?
b. What is the car's position after 5.00 seconds?
c. After the initial 5.00 seconds the car travels at a constant speed you found in (a). How long does it take for the car to travel 250 meters?

Expert's answer

(a) The speed of the car after 5 s


vf=vi+at=0+6.50 m/s2×5.0 s=32.5 m/sv_f=v_i+at=0+6.50\:\rm m/s^2\times 5.0\: s=32.5\: m/s

(b) The position of the car after 5 s


xf=xi+vit+at22=6.50 m/s2×(5.0 s)22=81.25 mx_f=x_i+v_it+\frac{at^2}{2}=\frac{6.50\:\rm m/s^2\times (5.0\:\rm s)^2}{2}=81.25\:\rm m

(c) The time needed to travel 250 m with velocity 32.5 m/s


t=dv=250 m32.5 m/s=7.7 st=\frac{d}{v}=\frac{250\:\rm m}{32.5 \:\rm m/s}=7.7\:\rm s


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