## 1. What is acceleration?

**Answer**: Acceleration is the rate of change of velocity or it can be simply defined as change in velocity per time. It is given mathematically as

Where a = acceleration, V=velocity and t= time taken

### 2. What is S.I unit of acceleration?

**Answer**: It is m/s^{2} (that is meter per a Second Square)

### 3. Define uniform accelerations

**Answer**: Acceleration is said to be uniform acceleration when the rate of change of velocity with time is constant. That is

###

4. What is retardation?

**Answer:** Retardation or deceleration is when
the velocity of a body is decreasing with time. In physics it can also be
called negative acceleration.

###

5. List four equation of uniformly accelerated motion

**Answer:**

##

6. What quantity is acceleration? Scalar or vector.

**Answer:** Acceleration is a vector quantity,
because it has both magnitude and direction.

####

7. A car starts from the rest and accelerated uniformly until it reaches a velocity of 30m/s after 5s. It travels with uniform velocity for 15s and is then brought to rest in 10s with a uniform retardation. (a). Draw the velocity time diagram

(b).Calculate the acceleration

(C). Determine the retardation

**Answers**: (a)

(b) The acceleration = the slope of OA

That is

(c). the retardation =the slop of BC

That is

###

8. An object moves in a straight line, starting from rest. There are two stages in the journey.

(a) It gained speed uniformly for 2.0s and attained a speed of 8.0m/s

(b) It continues at this speed for 1.5s.

Draw a sketch-graph of speed against time and from the sketch determine

(i).The accelerations in stage (a)

(ii).the accelerations in stage (b)

**Answer**:

(i)The acceleration in stage (a) = slope of OA, that is

(ii) The acceleration at stage (b) =0, since there speed was not changing

####

9. A train starts from rest from a station and travels with uniform acceleration 0.5ms^{-2} for 20s. it travels with uniform velocity for another 30s, the brakes are then applied so that a uniform retardation is obtained and the train comes to rest in further 10s. Sketch the velocity time graph of this motion. Using your graph, calculate the total distance traveled by the train.

**Answer:**

where a=0.5, t=20s therefore V=0.5 x 20 =10m/s

To calculate the total area covered by the train, you calculate the area of the diagram above, which is the area of the triangle OAE + the area of the rectangle ABDE + the area of the triangle BCD, so the total distance covered is

½ (20 x 10) + (10 x30) + ½(10 x 10)

½ (200) + 300 + ½(100)

100 + 300 +50 = 450m

So the distance covered by the train =450m

###

10. A motor car is uniformly retarded and brought to rest from a velocity 36km/h in 5s. Find it retardation and the distance covered during the period.

**Answer:** Note that retardation is also known as negative acceleration; therefore it can be solved using

Where v=36km/h=10/s, and t =5s, therefore

-a= 10/5= -2m/s^{2}

For the distance covered we use

**V ^{2}=U^{2}
+ 2aS**, Where v=final velocity=10m/s, U=initial velocity = 0,
-a =retardation =-2m/s2 and S =distance covered =? Therefore

10^{2} =0^{2} + 2(-2 x S)

100 = 0 -4S

S= 25m

So the distance covered by the motor =25m

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