Photo AI

Use the formula $s = ut + \frac{1}{2}at^2.$ (a) Calculate $s$ when $u = 5$, $t = 10$ and $a = 3.$ (b) Make $a$ the subject of the formula. - OCR - GCSE Maths - Question 1 - 2017 - Paper 1

Question icon

Question 1

Use-the-formula-$s-=-ut-+-\frac{1}{2}at^2.$--(a)-Calculate-$s$-when-$u-=-5$,-$t-=-10$-and-$a-=-3.$--(b)-Make-$a$-the-subject-of-the-formula.-OCR-GCSE Maths-Question 1-2017-Paper 1.png

Use the formula $s = ut + \frac{1}{2}at^2.$ (a) Calculate $s$ when $u = 5$, $t = 10$ and $a = 3.$ (b) Make $a$ the subject of the formula.

Worked Solution & Example Answer:Use the formula $s = ut + \frac{1}{2}at^2.$ (a) Calculate $s$ when $u = 5$, $t = 10$ and $a = 3.$ (b) Make $a$ the subject of the formula. - OCR - GCSE Maths - Question 1 - 2017 - Paper 1

Step 1

Calculate $s$ when $u = 5$, $t = 10$ and $a = 3$

96%

114 rated

Answer

To calculate ss, we will use the given values in the formula:

s=ut+12at2s = ut + \frac{1}{2}at^2

Substituting the values:

  • u=5u = 5
  • t=10t = 10
  • a=3a = 3

We can substitute these values into the equation:

s=(5)(10)+12(3)(102)s = (5)(10) + \frac{1}{2}(3)(10^2)

Calculating each part:

  1. Calculate utut:
    5×10=505 \times 10 = 50
  2. Calculate 12at2\frac{1}{2}at^2:
    12(3)(100)=150\frac{1}{2}(3)(100) = 150

Adding these results together: s=50+150=200s = 50 + 150 = 200

Thus, the calculated value of ss is 200.

Step 2

Make $a$ the subject of the formula

99%

104 rated

Answer

Starting with the original formula:

s=ut+12at2s = ut + \frac{1}{2}at^2

We want to isolate aa. First, we can rearrange the equation:

  1. Subtract utut from both sides:
    sut=12at2s - ut = \frac{1}{2}at^2
  2. Multiply both sides by 2:
    2(sut)=at22(s - ut) = at^2
  3. Finally, divide both sides by t2t^2 to solve for aa:
    a=2(sut)t2a = \frac{2(s - ut)}{t^2}

Now, aa is isolated as the subject of the formula.

Join the GCSE students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;