Question

A powerful motorcycle can accelerate from rest to 28m/s in only 4s . what is its average acceleration ? How far does it travel in that time.

Answer 1

Make use the following kinematics equations:

SUVAT-- s-displacement, u-initial velocity, v-final velocity,a- acceleration, t-timev=u + atv2 = u2 + 2ass=1/2(u+v)ts=ut + 1/2at2

The values for the motor cycle are as follows:
u= 0m/s
v=28m/s
t=4s
a=??s=??
v=u +at
28= 0 + 4a
4a=28                            a= 7m/s2

Therefore the average acceleration = 7m/s2

displacement s,  v2=u2+2as
                          28^2= 0^2+(2x7s)
                         784= 14s
                         s= 784/14
                          s= 56
Therefore the distance travelled = 56 meters.

Answer 2

Answer:

SUVAT-- s-displacement, u-initial velocity, v-final velocity,a- acceleration, t-timev=u + atv2 = u2 + 2ass=1/2(u+v)ts=ut + 1/2at2

Explanation:

The values for the motor cycle are as follows:

• u= 0m/s
• v=28m/s
• t=4s
• a=??s=??
• v=u +at
• 28= 0 + 4a
• 4a=28                            a= 7m/s2

1. Therefore the average acceleration = 7m/s2

displacement s,  v2=u2+2as

                         28^2= 0^2+(2x7s)

                        784= 14s

                        s= 784/14

                         s= 56

2.Therefore the distance travelled = 56 meters.

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