Question

A soccer ball is moving horizontally at a speed of 4.0 m/s. It then undergoes a constant
acceleration. After 8.00 s, the ball is moving at 4.8 m/s. What is the ball’s displacement? aa

Answer 1

Answer:

35.2 m

Explanation:

We have the initial and final velocities.

V1=4 m/s, V2=4.8 m/s

After 8.00 s

so now acceleration would be

a=(V2-V1)/time.

a=(4.8-4)/8

a=0.1 m/s^2

we know that displacement 's' is

s= ut+1/2at^2

so s= 4*8+ 1/2(0.1*8^2)

By solving we get S= 35.2m.

Answer 2

$initial \: speed \: of \: ball \: = 4 \: \frac{m}{s} \\ \\ time \: in \: constant \: acceleration \: = 8 \: sec \\ \\ final \: speed \: of \: ball \: = 4.8 \: \frac{m}{s} \\ \\ \huge \maltese \: \green{accleartion \: of \: ball \: } \\ \\ using \: first \: equation \: of \: motion \: v \: = \: u + at \\ \\ 4.8 = 4 + 8a \\ \\ a = \frac{.8}{8} \\ \\ a = 0.1 \: \frac{m}{s^2} \\ \\ now \: using \: the \: formula \: \: \: s = \frac{v {}^{2} - u {}^{2} }{2a} \\ \\ s \: = \frac{(4.8 ){}^{2} - (4) {}^{2} }{0.2} \\ \\ s = \frac{7.04}{2} \\ \\ s = 3.52 \: m$