Question

a powerfull motorcycle can accelerate from rest to 28m/s in only 4sec(a) what is its average acceleration. (b) how far does it travel in that time ?

Answer 1

Answer:

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.

Explanation:

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Answer 2

Answer:

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Explanation:

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.