First: Without the proper tools you're doomed to fail, so at the very least do yourself a favor and get yourself a digital multimeter and learn how to use it.
Second: When you ask a question you should provide at least enough information (Like a schematic diagram of your circuit) to show how you connected all of the circuit elements together.
If you don't do that, then how are we supposed to know whether you connected the power supply backwards or made some other gross error which let the magic smoke escape from the circuit?
The short answer is, "We won't", and all we'll wind up doing is wasting a lot of time guessing at what your problem might be before your question gets closed.
The drawing below shows your circuit, which is an emitter follower, and a common emitter circuit.
The reason the emitter follower circuit doesn't work very well for your application is because the base-to-emitter junction looks like a diode and needs to drop about 0.7 volts higher than the emitter voltage before the collector can start supplying current into the load connected to the emitter. However, with the motor in there, as the current through the motor starts to increase, its voltage drop increases and starts pushing back on the emitter, so that the base-to-emitter voltage drops and the current into the load starts being limited. In your case, since there's only 3 volts available to drive the base and the base-to-emitter diode takes about 0.7 volts of that, the highest voltage that can be developed across the motor is 3 volts - 0.7 volts, which is 2.3 volts, and that's why your motor doesn't work very well.
In the common emitter circuit, however, the motor is located between the supply and the collector, so its voltage drop doesn't affect the emitter much, its being connected directly to the negative (ground) side of the supply.
That being the case, all that's necessary to turn the transistor on and put nearly all of the 5 volts across the motor is to put enough current into the base.
A transistor has what's called "current gain", or "beta", which means that if it has a beta of 100 and I want one ampere of collector current, what I have to do is force 10 milliamperes of current into the base and that'll turn the transistor ON enough to let 1 ampere flow from the supply through the load, then through the transistor's collector-to emitter junction, and then from the emitter back to the supply.
In switching situations, though, what we do to make sure that the transistor is fully turned on is to push enough current into the base to saturate the collector-to emitter junction by making the base current about 10% of the load current.
In the example I've given, I've assumed a load current of about 500 milliamperes for the motor and, consequently, forced 50 milliamperes into the base for what's called a "forced beta" of 10. If the load current is less than 500 milliamperes it won't matter at all because the transistor will still be fully turned on, but if it's a lot more than 500 milliamperes, then the transistor's data sheet will need to be reviewed to make sure everything's OK.
The diode across the motor is used to clamp the high voltage spike which will occur when the motor is abruptly turned off - and would otherwise destroy the transistor - to Vcc + 1 diode drop, or about 6 volts.