a motorcycle can accelerate from rest to 28m/s in only 4s calculate (a) acceleration, (b) how far does it travel in that time
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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/sv=28m/st=4sa=??s=??v=u +at28= 0 + 4a4a=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= 56Therefore the distance travelled = 56 meters.
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/sv=28m/st=4sa=??s=??v=u +at28= 0 + 4a4a=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= 56Therefore the distance travelled = 56 meters.
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