Question

A body starts from rest and moves with a uniform acceleration of 10ms^-2 in the first 10 seconds. During the next 10 seconds. It moves with the maximum velocity attained. The total displacement of the body is

Answer 1

The initial velocity is u=0ms−1

The acceleration is a=10ms−2

The time is t=5s. Apply the equation of motionv=u+at

The velocity isv=0+10⋅5=50ms−1

The distance travelled in the first 5s is s=ut+12at2=0⋅5+12⋅10⋅52=125m

The distance travelled in the following 10s iss1=vt1=50⋅ 10=500m

The total distance travelled is d=s+s1=125+500=

625m

Answer 2

Using above formula,

$d1 = 0 + \frac{1}{2} a {t}^{2}$

$d1 = \frac{1}{2} (10){10}^{2}$

$d1 = 500 \: m$

Using second formula,

$v(max) \: = 0 + at$

$v(max)= 0 + (10)(10)$

$v(max) = 100 \: \frac{m}{s}$

Now,

$d2 = v(max) \: t$

$d2 = (100)(10)$

$d2 = 1000 \: m$

Total distance = d1 + d2
Total distance = 500 + 1000
Total distance = 1500 metres