An electric toy car with a mass of #3 kg# is powered by a motor with a voltage of #8 V# and a current supply of #3 A#. How long will it take for the toy car to accelerate from rest to #4 m/s#?

2 Answers
Feb 22, 2017

The time is 1 second.

Explanation:

Given:
#mass = 3" kg"#
#V_"motor" = 8" Volts"#
#I_"motor" = 3" Amperes"#
#Velocity = 4" m/s"#

Let t = the time of acceleration.
Let #P =# the power of motor #= (V_"motor")(I_"motor")#

The energy provided by the motor is:

#E_"motor" = Pt#

The kinetic energy of the vehicle:

#E_"kinetic" = 1/2mass(Velocity)^2#

The following step assumes no loss in the system:

#E_"motor" = E_"kinetic"#

#Pt = 1/2mass(Velocity)^2#

#t = (mass(Velocity)^2)/(2P)#

#t = (3" kg"(4" m/s")^2)/(2(8" Volts")(3" Amperes")#

#t = (3" kg"(16" m"^2"/s"^2))/(2(24" Watts")#

#t = 1" s"#

Feb 22, 2017

The time is #=1s#

Explanation:

The power is

#P=UI#

#U=8V#

#I=3A#

#P=3*8=24W#

The change in kinetic energy is

#DeltaKE=1/2mv^2-1/2m u^2#

#u=0#

#v=4ms^-1#

#m=3kg#

#DeltaKE=1/2*3*16-0#

#=24J#

But

#P=(DeltaKE)/t#

So,

#t=(DeltaKE)/P#

#=24/24=1s#